NCERT Solutions Class 9 Science Chapter 7 Work Energy and Simple Machines

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

CLASS 9 · SCIENCE

NCERT Solutions

Chapter 7: Work, Energy, and
Simple Machines

NCERT Textbook — Exploration

BOOK PAGES SECTIONS QUESTIONS MEDIUM

116 – 139 21 47 English

Solutions, notes, sample papers & more at 63 pages

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

CLASS 9 · SCIENCE · EXPLORATION

NCERT Solutions — Chapter 7: Work, Energy, and
Simple Machines
Newton's laws tell you what a force does moment by moment. This chapter gives you a shortcut: keep track of
energy instead. Work done on an object equals the change in its energy — and from that one statement
come the formulae for kinetic energy, potential energy, the conservation of mechanical energy, power, and
the mechanical advantage of a pulley, an inclined plane and a lever.

TEXTBOOK BOOK PAGES

Exploration (Class 9) 116 – 139

SECTIONS QUESTIONS

21 47

MEDIUM

English

Think It Over — Page 116

Page 1 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Chapter opener

THINK IT OVER

Q1 What will be the magnitude of velocity of the child at the bottom of the blue slide?

Blue slide (spiral) Platform

Purple slide (wavy)
h
Red slide (straight)

Ground

Chapter opening picture, page 116 — redrawn sketch. Three slides of different shapes —
a blue spiral one, a straight red one and a wavy purple one — all start from the same
platform at height h above the ground.

If h is the height of the platform above the ground, the child arrives at the bottom with speed v
= √(2gh).

At the top: kinetic energy = 0, potential energy = mgh

At the bottom: potential energy = 0, kinetic energy = ½mv2

Conservation of mechanical energy (ignoring friction):

½mv2 = mgh

v2 = 2gh

v = √(2gh)

For the playground in the picture the deck is about 2 m above the ground, so

Page 2 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

v = √(2 × 10 m s–2 × 2 m) = √(40 m2 s–2) ≈ 6.3 m s–1

Why it happens: the mass m cancels from both sides, and so does the shape of the
slide. Only the vertical drop h decides the speed, because only the vertical drop
decides how much gravitational potential energy is converted into kinetic energy.

Q2 Will two children of different masses reach the bottom of the same slide with the
same velocity?

Yes — both reach the bottom with the same speed, √(2gh).

½mv2 = mgh

The mass m appears on both sides and cancels

v = √(2gh) — independent of m

Why it happens: a heavier child does store more potential energy at the top (mgh is
larger), but that same larger mass has to be accelerated. The two effects cancel
exactly, just as all bodies fall freely with the same acceleration g.

Tip: in real life friction between the child's clothes and the slide does work, and that
work depends on how hard the child presses on the surface. So a very light child on
a slow, rough slide can arrive a little slower than this ideal answer.

Q3 Which of the slides will result in the largest magnitude of velocity for the child at its
bottom?

Look at the picture on page 116: the blue spiral slide, the red straight slide and the purple wavy
slide all begin at the same platform and end at the same ground. So the vertical drop h is the
same for all three, and — ignoring friction — all three give the same speed at the bottom, v =
√(2gh).

Page 3 of 63

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Class 9 Science Chapter 7 Work, Energy, and Simple Machines
g l AglaSem · NCERT Solutions

co m
e m.
v = √(2gh) depends only on h
m l as
.co
same h → same v, whatever the length or shape of the path
m a g
l a se
g
aWhy it happens: gravity does work only through the vertical drop. Sliding sideways

com
or curling round a spiral adds path length but no extra height, so it adds no extra
energy.
e m . ag
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Check it yourself: once friction is allowed for, the long blue spiral rubs the child over

. c om
a much greater length of surface, so it takes away the most energy. In practice the
m would
short steep red slide gives the fastest arrival. A slide whose top was ehigher
s
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three.
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In-text Questions — Page 116
Chapter opener
se m
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But when forces change with time or act in complicated ways, applying these laws
Q1
directly can become difficult. Is there a simpler and more powerful way to
co m
m .
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understand such situations?
m l as
.co a g
a s em

agl Yes — keep track of energy instead of force. The work–energy theorem compares only the
starting and finishing states of an object, so you never have to follow the force instant by
se m
instant.
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m . a
ase
agl
Newton's laws need the force at every moment: F = ma gives a, then the kinematic equations
give v and s. If F keeps changing, a keeps changing and the kinematic equations no longer
apply.

co
The energy method needs only two numbers: the energy at the start and the energy at the m
m .
as e
end. Work done = change in energy holds whether the force is constant or not.
m l
.co a g
m Why it helps: in Example 7.9 a runaway truck rolls up a sand-filled escape ramp. The
s e
agla force from the sand, the slope and gravity all act together, but writing "initial energy
.c
+ work done by sand = final energy" gives the ramp length in three lines. Doing the
s e m
m a
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same problem with F = ma would take a page.
m .
as e
a g l
Pause and Ponder — Page 119
co m
m .
m ase
.co


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Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

7.1 Work Done by a Constant Force

PAUSE AND PONDER

Q1 In the previous chapter, a weightlifter is shown holding a barbell steady in her
hands (Fig. 6.8). Is she doing any work on the barbell while holding it steady?

Barbell

Weightlifter

Floor

Fig. 6.8, page 97 (Chapter 6) — redrawn sketch: a weightlifter holding the barbell steady
above her head.

No. She does zero work on the barbell, because the barbell has no displacement.

W=F×s

The upward force she applies = weight of the barbell = mg

Displacement of the barbell, s = 0

W = mg × 0 = 0 J

Why it happens: in science, work needs a displacement in the direction of the force.
The barbell stays where it is, so no energy is transferred to it — its kinetic energy
stays zero and its potential energy stays the same.

Page 5 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

But she does get tired. To keep applying the force, the fibres in her muscles
repeatedly contract and relax. That uses up the chemical energy stored in her body
and turns much of it into heat. So the energy is spent inside her body, not on the
barbell. Feeling tired is not the scientific test for work being done.

Q2 Is the work done by friction on the stack of coins that travels on a rough surface
(Fig. 6.13c) — positive, negative or zero?

Force of friction

A B
Force due to rubber band

(a)

A B

Force of friction

(b)

Fig. 6.13, page 99 (Chapter 6): forces acting on the stack of coins — (a) due to the
stretched rubber band and friction on release of the rubber band, and (b) due to friction
alone, once the coins have left the band and are sliding on the rough surface. (The book
prints only parts (a) and (b); the question’s “Fig. 6.13c” is the situation of part (b).)

The work done by friction on the stack of coins is negative.

Page 6 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Direction of motion of the coins →

Direction of the force of friction ← (always opposes relative motion)

Force and displacement are opposite
W = F × (– s) = negative

Why it happens: friction acts backwards on the sliding coins, so it takes energy
away from them instead of giving it. That lost kinetic energy appears as heat in the
coins and in the table top — which is exactly why the stack slows down and finally
stops.

Check it yourself: rub your palms together quickly. They get warm. The negative
work done by friction on your moving palms has become thermal energy.

In-text Questions — Page 120
7.2 The Work-Energy Theorem

Q1 We have seen that when a force is applied and an object is displaced, work is done
on it. Does this cause the object to gain capacity to do further work?

Yes. Positive work done on an object gives it energy, and energy is the capacity to do work.

A fielder does work on a cricket ball while throwing it. The moving ball then knocks the
wicket over — it has done work on the wicket (Fig. 7.8a, b).
Someone does work in raising a flowerpot to a window ledge. If the pot falls, it can damage
what lies below — again it does work (Fig. 7.8c).

work done on the object = change in its energy (Eq. 7.3)

positive work → energy gained → capacity to do work on something else

Why it happens: energy is not used up in the transfer — it moves on. The ball keeps
the energy the fielder gave it until it meets the wicket, and then hands it over. When
the ball does positive work on the wicket, the wicket does an equal negative work on
the ball, so the ball loses exactly what the wicket gains.

Page 7 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Pause and Ponder — Page 121
7.2 The Work-Energy Theorem

PAUSE AND PONDER

Q3 When you pedal a bicycle on a flat road, your muscles supply energy. In what forms
does this muscular energy appear as you ride?

The chemical energy in your muscles is converted into these forms:

Kinetic energy of the bicycle and of yourself — while you are speeding up. Once you ride at
a steady speed this stops growing.
Thermal energy — the largest share. Work done against friction at the axles, the chain and
the tyre–road contact, and against air resistance, heats the parts, the road surface and the
air.
Sound energy — the small noise of the chain, the tyres and the wind.
Thermal energy inside your body — muscles are not perfectly efficient, so you warm up
and sweat.

On a flat road there is no gain in potential energy (h does not change)

At a steady speed there is no gain in kinetic energy either

So: energy supplied by muscles = work done against friction and air resistance

= thermal energy + a little sound energy

Why it happens: at a steady speed the net force on you is zero, so your forward
push exactly balances friction and drag. Every joule your muscles put in is
immediately carried away by those opposing forces, which do an equal amount of
negative work. Stop pedalling and the bicycle coasts to a halt, because the drain
continues while the supply has stopped.

In-text Questions — Page 121

Page 8 of 63

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Class 9 Science Chapter 7 Work, Energy, and Simple Machines
g l AglaSem · NCERT Solutions

7.4.1 Kinetic energy
co m
e m.
m l as
Q1
.co
How much is the energy possessed by an object by virtue of its motion?
m a g
l a se
a g

An object of mass m moving with speed v has kinetic energy K = ½mv2.
com
e m . ag
g l as
a
Take an object of mass m pushed by a constant force F from speed u to speed v over

displacement s.

co m
m.
Kinematics: v2 = u2 + 2as, so s = (v2 – u2) / 2a
m as e
.co
Work done: W = F × s = ma × (v2 – u2)/2a
a g l
a s e=m½m(v2 – u2) (Eq. 7.5)
a gl W
By the work–energy theorem this is the change in the object's energy.
m a s
.co
Starting from rest (u = 0), the whole of it is the kinetic energy gained:
m agl
l a se
K = ½mv2 (Eq. 7.6)
a g

co m
Why the square matters: K grows as v2, not as v. Doubling the speed of a vehicle
m .
o m l a se
makes its kinetic energy four times larger (Example 7.4), so the brakes must remove
g is the biggest factor in
.c times as much energy — which is why speed, not mass,
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aseroad accidents.
four

agl
se m
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.
Tip: kinetic energy has no direction. A ball moving east at 5 m s–1 and the same ball
m a
ase
moving west at 5 m s–1 have exactly the same kinetic energy. Its unit is the joule: 1 J
= 1 kg m2 s–2. agl

co m
m .
as e
omand Ponder — Page 123
Pause
. c a g l
s e m
a
agl c
m .
m a s e
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g l as
a

com
m .
m ase
.co


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Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

7.4.1 Kinetic energy

PAUSE AND PONDER

Q4 Two objects A and B of mass m and 4 m have the same kinetic energy. What is the
ratio of the magnitude of velocities of A and B?

The ratio is vA : vB = 2 : 1.

K = ½mv2
Given KA = KB

½ × m × vA2 = ½ × 4m × vB2

m cancels: vA2 = 4 vB2

Taking the square root: vA = 2 vB

vA : vB = 2 : 1

Why it happens: kinetic energy depends on mass to the first power but on speed to
the second power. To make up for being 4 times lighter, A needs to be only √4 = 2
times faster.

Check it yourself: put numbers in. Let m = 1 kg and vB = 3 m s–1. Then KB = ½ × 4 kg
× 9 m2 s–2 = 18 J, and KA = ½ × 1 kg × 62 m2 s–2 = 18 J. They match.

Q5 Does the kinetic energy of an object which moves with constant velocity change
with its position?

No. Its kinetic energy stays exactly the same wherever it is.

K = ½mv2

The formula contains only m and v — no h, no x, no position of any kind.

Constant velocity → v does not change → K does not change

Page 10 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Why it happens: constant velocity means zero acceleration, so by Newton's first law
the net force on the object is zero. Zero net force does zero net work, and by the
work–energy theorem zero work means zero change in energy.

Careful: its potential energy can still change if the position change is vertical — but
that would need some other force to lift it, and then the motion would not stay at
constant velocity under gravity alone. For horizontal motion at constant velocity,
both K and U stay unchanged.

Activity 7.1: Let us investigate — Page 125
7.4.2 Potential energy — Gravitational Potential Energy

ACTIVITY

Q1 Raise the ball over the sand bed to a height of about 1 m and drop it (Fig. 7.17). Is a
depression created in the sand? Why does the ball create a depression?

ball dropped from a smaller height ball dropped from a greater height

Loose sand

Container / tray

Fig. 7.17, page 125 — redrawn sketch: depressions created in loose sand by a ball
dropped from different heights.

Yes, a bowl-shaped depression forms. The ball creates it because it arrives with kinetic energy
and uses that energy to do work on the sand, pushing the grains aside.

Page 11 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Raising the ball through h = 1 m stores potential energy U = mgh

Falling freely, all of it becomes kinetic energy just above the sand:

½mv2 = mgh, so v = √(2gh) = √(2 × 10 × 1) ≈ 4.5 m s–1

In the sand, the resistive force F acts through depth d and stops the ball:

F × d = mgh → d = mgh / F

Why it happens: the sand pushes back on the ball (negative work on the ball) and
the ball pushes the grains outward and downward (positive work on the sand). The
ball keeps moving until the sand has absorbed all its kinetic energy — and the
deeper it goes, the more work the sand has done.

Try This: repeat on a hard cemented floor. There is no depression at all, because the
floor cannot be deformed — the energy goes instead into a loud sound and into the
bounce of the ball.

Q2 Now, raise the ball to the height of 2 m and release it at a slightly different position
over the sand bed such that the depressions do not overlap. Repeat this step one
more time. Compare the depths of the depressions. Is there any difference? In
which case is the depression deepest and in which case the shallowest?

Yes, there is a clear difference. The depression made by the ball dropped from 2 m is the
deeper one; the one from 1 m is the shallowest. Dropping again from 2 m gives a depression
of about the same depth as the first 2 m drop.

HEIGHT OF DROP, POTENTIAL ENERGY STORED, U = DEPTH OF DEPRESSION, D =
H MGH MGH/F

1m mg × 1 m shallowest

2m mg × 2 m — twice as much about twice as deep

Page 12 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Why it happens: raising the ball to a greater height means doing more work against
gravity, so more potential energy is stored. All of it arrives as kinetic energy at the
sand, and the sand must do a matching amount of negative work to stop the ball.
With the same average resistive force F, twice the energy means twice the stopping
distance — a deeper pit.

Conclusion of the activity: the greater the height of an object above the Earth's
surface, the greater its gravitational potential energy — which is the result the book
goes on to write as U = mgh (Eq. 7.8).

Pause and Ponder — Page 126
7.4.2 Potential energy

PAUSE AND PONDER

Q6 Does the potential energy of an object near the surface of the Earth change if it
moves with constant velocity in the horizontal direction? What if the object is
gradually raised in the vertical direction?

Horizontal motion: no change. Vertical raising: the potential energy increases by mgΔh.

U = mgh

(a) Horizontal motion: the height h above the ground does not change

ΔU = mg × 0 = 0 J — no change

(b) Raised gradually through Δh:

ΔU = mg × Δh — the potential energy increases

e.g. a 2 kg book lifted 1.5 m: ΔU = 2 kg × 10 m s–2 × 1.5 m = 30 J

Why it happens: gravity pulls straight down. Moving sideways is perpendicular to
that pull, so gravity does no work and nothing is stored. Moving up is directly against
the pull, so you must do work mgΔh against gravity — and that work is stored in the
Earth–object system as potential energy, ready to be returned if the object is
released.

Page 13 of 63

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Class 9 Science Chapter 7 Work, Energy, and Simple Machines
g l AglaSem · NCERT Solutions

co m
m.
Tip: "gradually" matters. If you raise the object slowly the applied force is just mg

m as e
l
and there is no leftover kinetic energy — every joule you supply goes into potential
energy. .co a g
a s em
agl
om127
Activity 7.2: Let us experiment — Page
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7.4.3 Conservation of mechanical energy

ACTIVITY agl

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Take the bob to one side to a point P, which is at the level of the horizontal line and

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Q1

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let it go. Observe it at the extreme points of the first couple of oscillations. Does the
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bob reach the level of the horizontal line?

la s
agANSWER
a s
comfurther short with every swing. agl
Yes — on the first few swings the bob rises to almost the horizontal line on the far side,
but never quite reaches it, and it falls a .little

a s em
POINT HEIGHT ABOVE ag
l POTENTIAL KINETIC MECHANICAL
LOWEST POINT ENERGY ENERGY ENERGY

. com
e m
as
P h mgh 0 mgh

m l
.co g
(release)

eQm a
a s
gl (lowest)
0 0 ½mv2 = mgh mgh

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se m
R (far h (almost) mgh
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0 mgh
g l a
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side)

a gl
Why it happens: the string is inextensible and the tension in it is always
perpendicular to the bob's motion, so the string does no work. Only gravity does
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work, and gravity simply trades potential energy for kinetic energy and back again.
m l as
.co g
The sum K + U — the mechanical energy — stays at mgh, so the bob must return to

em a
s
the height h it started from.
l a
ag c
m .
Why "almost": in the real world friction at the support and air resistance do a small
a s e
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amount of negative work on every swing. That energy leaves as heat and sound, so
agl
a s emfalls, the bob rises a little less each time, and the
the mechanical energy slowly

agl
pendulum finally stops.

co m
m .
m as e
.co


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Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Pause and Ponder — Page 129
7.4.3 Conservation of mechanical energy

PAUSE AND PONDER

Q7 For the situation depicted in Fig. 7.19, calculate the mechanical energy of the ball
just before it hits the ground and show that even at this position, it is mgh.

A t=0

mg
(h − h′)

h B t=t
m

h′

C

Fig. 7.19, page 126: an object of mass m dropped from rest at A (height h), passing B
(height h′) after time t, and reaching the ground at C.

Just before it reaches C the ball has no potential energy left and kinetic energy mgh, so its
mechanical energy is still mgh.

Page 15 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

The object is released from rest at A, at height h. Take the ground (C) as h = 0.

Height at C: h′ = 0, so potential energy = mg × 0 = 0 J

Speed at C — using v2 = u2 + 2gh with u = 0:

v2 = 0 + 2gh = 2gh

Kinetic energy at C:

K = ½mv2 = ½ × m × 2gh = mgh

Mechanical energy at C = K + U = mgh + 0 = mgh ✔

Compare this with the two positions the chapter already worked out:

POINT POTENTIAL ENERGY KINETIC ENERGY MECHANICAL ENERGY

A (t = 0, height h) mgh 0 mgh

B (time t, height h – ½gt2) mgh – ½mg2t2 ½mg2t2 mgh

C (ground, height 0) 0 mgh mgh

Why it happens: gravity is the only force doing work here. Whatever potential
energy the ball loses by falling, it gains back exactly as kinetic energy. The store
simply changes its label, so the total never moves off mgh. This is the conservation
of mechanical energy.

Check it yourself: take m = 0.5 kg, h = 5 m, g = 10 m s–2. At A: U = 0.5 × 10 × 5 = 25 J.
At C: v = √(2 × 10 × 5) = 10 m s–1, so K = ½ × 0.5 × 100 = 25 J. Same number.

Page 16 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Q8 You may have seen an exhibit like that in Fig. 7.22 in a science park, where a ball is
released from the highest point. Describe how the kinetic energy and potential
energy change at points A, B and C. Why do subsequent points, such as C, D and E,
usually have lower heights compared to the previous ones? Could it have anything
to do with the energy lost due to friction?

ball released here

A

C

D
E

B

Base

Fig. 7.22, page 129 — redrawn sketch of the ball roller-coaster exhibit: the ball is
released at the top of the tall tower, rises to peak A, dips to B, and then over the
successively lower peaks C, D and E.

The ball is released from rest at the top of the tall tower, so its whole mechanical energy is
potential energy there. After that the two forms keep swapping — but the total slowly leaks
away.

Start starting height — never reached again

A
C

D

E

B
direction of travel

Page 17 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Each hump is lower than the one before because friction and air resistance remove a little mechanical
energy on every stretch of track.

POINT HEIGHT POTENTIAL KINETIC ENERGY
ENERGY

A — first hump high, but below the large small (only the part of the drop
start already made)

B — valley between lowest, about ground almost zero — minimum maximum — the ball is fastest here
humps level

C — next hump lower than A large again, but less than more than at A
at A

At every point: K + U = mechanical energy

Going down a slope: U falls, K rises by the same amount
Going up a hump: K falls, U rises by the same amount

Why C, D and E are lower each time — yes, friction is exactly the reason.

Height a ball can climb, h = mechanical energy ÷ mg

On each stretch of track, friction and air resistance do negative work

mechanical energy after = mechanical energy before – (energy lost to friction)

Less mechanical energy → smaller maximum height on the next hump

Why it happens: friction between the ball and the rail, and air resistance, always act
opposite to the ball's motion. That negative work converts a little mechanical energy
into heat and sound on every pass, and heat and sound cannot climb back into the
ball. So the designer must make every hump lower than the last — otherwise the
ball would not have enough energy left to get over it and would roll back.

Did you know? Real roller coasters are built the same way. The first drop is always
the tallest, and every loop and hill after it is shorter than the one before.

In-text Questions — Page 131

Page 18 of 63

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Class 9 Science Chapter 7 Work, Energy, and Simple Machines
g l AglaSem · NCERT Solutions

7.6.2 Inclined plane
co m
em.
m l as
Q1
.co a
You can place the box on a smooth inclined plank as shown in (Fig. 7.26b), and push
m
g
l a se
the box up the plank to the platform. Does this method require a smaller force?

a g

co m
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g l as
F
a
inclined plank

co m
e m.
m l as
.co
F′
a g
se m
g l a
a (a) (b)

m a s
.co agl
Fig. 7.26, page 131 — redrawn sketch: a box being (a) lifted vertically up onto the

m
platform, and (b) pushed up a ramp to the same platform.

l a se
a g

co m

m .
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Yes. Pushing the box up a smooth ramp needs a smaller force than lifting it straight up —
m l as
.co g
smaller by the factor h/L.

em a
a s
agl Lifting straight up: force needed = weight = mg, distance = h
se m
com a
Pushing up a ramp of length L to the same height h at constant speed:
. a g l
m
ase
work you do = F′ × L

potential energy gained = mgh
agl
Work–energy theorem (ignoring friction): F′ × L = mgh
co m
m .
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F′ = mgh / L = mg × (h/L)
m l as
m .co
Since L > h, the factor h/L is less than 1, so F′ < mg
a g
l a se
ag
.c
For example, raising a 100 kg crate through h = 1 m using a ramp of length L = 4 m:

s e m
m a
e m . co
Lifting directly: F = mg = 100 kg × 10 m s–2 = 1000 N agl
g l as
a
Up the ramp: F′ = 1000 N × (1 m / 4 m) = 250 N — only a quarter as much

co m
m .
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.co


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Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Why it is not "free": the work done is the same both ways — 1000 J in this example.
On the ramp you apply 250 N over 4 m instead of 1000 N over 1 m. The machine
spreads the same work over a longer distance so that the force at any instant is
small enough for you to manage.

Activity 7.3: Let us experiment — Page 131
7.6.2 Inclined plane

ACTIVITY

Q1 Next, place the plank against the top of the pile of books or stool as shown in Fig.
7.27a. Pull the cart along the plank slowly and steadily. Is the reading of the spring
balance (the force required) smaller than that of step 2?

Spring balance
Cart

Plank

Pile of books
(a) (b)

Fig. 7.27, page 131 — redrawn sketch: measuring the force needed to pull a cart up an
inclined plank of (a) smaller length (steeper) and (b) larger length (less steep), the top of
the pile of books being at the same height in both.

Yes — the spring balance reads clearly less on the ramp than when the cart was lifted
straight up.

Page 20 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

HOW THE CART IS SPRING BALANCE DISTANCE WORK DONE
RAISED READS MOVED

Step 2 — lifted vertically mg (full weight) 0.5 m mg × 0.5 m
through h = 0.5 m

Step 3 — pulled up a plank of mg × (h/L) = mg/3, i.e. 1.5 m (mg/3) × 1.5 m = mg ×
length L = 1.5 m about one-third 0.5 m — same

Mechanical advantage of the plank = L / h = 1.5 m / 0.5 m = 3

So the effort is one-third of the weight.
If the cart weighs 6 N, the balance should read about 2 N on the ramp.

Why it happens: the plank supports part of the cart's weight through the normal
force. You only have to overcome the component of the weight that acts along the
slope, and the gentler the slope, the smaller that component is.

Check it yourself: measure the actual reading. It will be a little more than mg × h/L,
because friction between the cart's wheels and the plank has to be overcome too —
the derivation above ignores friction.

Page 21 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Q2 Now, reduce the angle between the plank and the base as shown in Fig. 7.27b, and
repeat the step 3. Observe how the force required changes as the plank becomes
less steep.

Spring balance
Cart

Plank

Pile of books
(a) (b)

Fig. 7.27, page 131 — redrawn sketch: measuring the force needed to pull a cart up an
inclined plank of (a) smaller length (steeper) and (b) larger length (less steep), the top of
the pile of books being at the same height in both.

The force required becomes still smaller as the plank is made less steep — but you must
now pull the cart over a longer distance.

For the same height h, a shallower plank means a longer L

Mechanical advantage = L / h → larger

Effort F′ = mg × (h / L) → smaller

Work done = F′ × L = mgh → unchanged

PLANK LENGTH L (H = MECHANICAL EFFORT FOR A 6 N WORK
0.5 M) ADVANTAGE L/H CART DONE

1.0 m 2 3.0 N 3.0 J

1.5 m 3 2.0 N 3.0 J

3.0 m 6 1.0 N 3.0 J

Page 22 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Why it happens: the inclined plane cannot reduce the work — that is fixed at mgh
by the height alone. What it can do is trade force for distance. Making the plank
twice as long halves the effort and doubles the distance, keeping the product exactly
the same.

Did you know? This is the whole idea behind the long, gently sloping ramps built for
wheelchairs, and behind hill roads that wind instead of climbing straight up.

Pause and Ponder — Page 132
7.6.2 Inclined plane

PAUSE AND PONDER

Q9 Explain why roads on hills are built to wind around in gentle slopes rather than
going straight up (Fig. 4.26)?

Mountain road

Fig. 4.26, page 67 (Chapter 4) — redrawn sketch: a mountain road that winds up the
hillside in gentle slopes and hairpin bends.

A winding hill road is a long inclined plane. It raises a vehicle through the same height h using
a much longer path L, so the force the engine must supply along the road is far smaller.

Page 23 of 63

as e
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Class 9 Science Chapter 7 Work, Energy, and Simple Machines
g l AglaSem · NCERT Solutions

co m
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For a slope, mechanical advantage = L / h (Eq. 7.13)
m l as
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Force needed along the road, F′ = mg × (h / L)
m a g
l a se
g
Winding the road makes L large for the same h → F′ becomes small
aWork done against gravity = F′ × L = mgh — the same either way

. c om ag
Suppose a 1500 kg truck must climb a hill e
s m m high, g = 10 m s–2:
a
300

agl
Work against gravity = mgh = 1500 kg × 10 m s–2 × 300 m = 4 500 000 J
co m
Straight up a 600 m ramp: F′ = 4 500 000 J ÷ 600 m = 7500 N
e m.
m l as
.co a g
Along a winding road 6000 m long: F′ = 4 500 000 J ÷ 6000 m = 750 N — one-tenth
m
l a se
a g
Why it happens: the road cannot reduce the energy the truck needs — that is mgh,
m a s
c o agl
fixed by the height of the hill. What the gentle slope does is spread that work over
.
ten times the distance, so the forcemneeded at any instant drops to one-tenth. An
a
ordinary engine and ordinaryltyresegrip can supply 750 N; they could not supply 7500
N. ag

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Two more reasons: on a gentle slope the tyres are less likely to e
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a g l Page 24 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Q10 To reach a higher floor, we find climbing an inclined ladder easier in comparison to
climbing a vertical ladder (Fig. 7.30). Explain why.

Inclined ladder Vertical ladder

Building

Fig. 7.30, page 132 — redrawn sketch: an inclined ladder and a vertical ladder set
against the same building to reach the same height.

The sloping ladder is an inclined plane. Each step lifts you through a smaller height, so each
push needs a smaller force — and the ladder itself supports part of your weight.

Vertical ladder: every step raises you the full rung spacing

→ each push must supply almost the whole weight mg, using your arms as well as your legs

Inclined ladder of length L reaching height h:
force along the ladder, F′ = mg × (h / L) < mg

Total work done = F′ × L = mgh — the same for both ladders

Why it happens: on a slanting ladder the rungs push back on your feet
perpendicular to the ladder, and this normal force carries a share of your weight.
You supply only the part of the weight acting along the slope, and you supply it over
a longer climb. On a vertical ladder the rungs push straight up but you must also
grip and pull to stay on it, so your arms do a large part of the lifting.

Page 25 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Tip: "easier" here means a smaller force at each moment, not less energy. Climbing
to a first floor 3 m up costs a 50 kg student mgh = 50 × 10 × 3 = 1500 J on either
ladder.

Activity 7.4: Let us investigate — Page 133
7.6.3 Lever

ACTIVITY

Q1 On the other end of the scale, place one eraser. Does the stapler lift up? If not, add
one more eraser.

With one or two erasers the far end of the scale goes down and the heavy stapler is lifted
— even though a stapler weighs several times as much as an eraser.

The scale is a lever, the pencil is the fulcrum.

Balance condition: effort × effort arm = load × load arm (Eq. 7.15)

Say the pencil is 5 cm from the stapler end and 25 cm from the eraser end:

load arm = 5 cm, effort arm = 25 cm

Mechanical advantage = effort arm / load arm = 25 cm / 5 cm = 5

So an eraser weighing 0.2 N can lift a stapler weighing up to 5 × 0.2 N = 1 N

Why it happens: the eraser end is far from the pencil, so it sweeps through a large
distance; the stapler end is close to the pencil and moves only a little. Work done at
one end is passed to the other end, F1d1 = F2d2. A small force moving far can
therefore deliver a large force moving a short way.

Try This: slide the pencil towards the middle of the scale. The mechanical advantage
falls, and now the erasers cannot lift the stapler. Slide it closer to the stapler and
even one eraser is enough. The lever's power lies entirely in where the fulcrum sits.

Page 26 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

In-text Questions — Page 133
7.6.3 Lever

Q1 By applying a small force at one end of the lever, a larger force can be applied on
the object on the other end of the lever. How is this possible?

It is possible because the lever passes on the work, not the force. The end you press moves
through a large distance with a small force; the load end moves a small distance with a large
force.

Load F₂ (large) Effort F₁ (small)

effort arm d₁ (long)

load arm d₂ (short)
Fulcrum

A small effort moving through a long arm delivers a large force through a short arm: F₁ × d₁ = F₂ × d₂.

Work put in at the effort end = work delivered at the load end

F1 × d1 = F2 × d2 (Eq. 7.14)

Rearranging: F2 = F1 × (d1 / d2)

Mechanical advantage = effort arm / load arm = d1 / d2 (Eq. 7.16)

Why it happens: the lever is a rigid bar turning about the fulcrum, so both ends
turn through the same angle in the same time. The end further from the fulcrum
must therefore travel a proportionally longer arc. Since the work done on the bar at
one end reappears at the other, a longer path must go with a smaller force, and a
shorter path with a larger one.

Page 27 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

What a lever cannot do: it does not reduce the total work, and it does not create
energy. Lift a 200 N rock by 0.1 m and you must supply 20 J, whether you use a
crowbar or your bare hands.

Activity 7.5: Let us experiment — Page 134
7.6.3 Lever

ACTIVITY

Q1 Record all observations and measurements, and complete the Table 7.1 by adding
more rows. [Number of coins in the left pan, n₁ (Effort); Distance of left pan from the
fulcrum, L₁ (cm); Number of coins in right pan, n₂ (Load); Distance of right pan from
the fulcrum, L₂ (cm)]

Table 7.1: Number of coins in the left pan and its distance from the fulcrum

NUMBER OF DISTANCE OF LEFT NUMBER OF DISTANCE OF
COINS IN LEFT PAN FROM THE COINS IN RIGHT RIGHT PAN FROM
PAN, N1 FULCRUM, L1 (CM) PAN, N2 (LOAD) THE FULCRUM, L 2
(EFFORT) (CM)

1 1

1 2

Table 7.1, page 134 — the two rows printed in the book. Copy it out and add more rows
for 4 coins and 8 coins in the right pan.

Keep the left pan (1 coin) at a fixed distance — say L1 = 20 cm — and slide the loaded right pan
until the beam is level each time. A typical completed table:

Page 28 of 63

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Class 9 Science Chapter 7 Work, Energy, and Simple Machines
g l AglaSem · NCERT Solutions

co m N₂ ×
mL₁.
COINS IN LEFT DISTANCE L₁ COINS IN RIGHT DISTANCE L₂ N₁ ×

a se
com
PAN, N₁ (EFFORT) (CM) PAN, N₂ (LOAD) (CM) L₂

. a g l
m
ase
1 20 1 20 20 20

a1gl 20 2 10 20 20

m
.co ag
1 20 4 5 20 20

a se8m
agl
1 20 2.5 20 20

co m
m.
The last two columns are always equal, so the beam balances when

m as e
.co l
n1 × L 1 = n 2 × L 2
a g
se m
a
that is, effort × effort arm = load × load arm (Eq. 7.15)

a g l

om a s
agl
Mechanical advantage = load / effort = n2/n1 = L1/L2

Row 4: MA = 8/1 = 20 cm / 2.5 cm =em
. c
a s 8

agl
Why it happens: the beam is a class-I lever with the string as fulcrum. Each coin is
co m
m .
e
an identical weight, so the number of coins measures the force directly. Doubling

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the load must halve its distance from the fulcrum, because it is the product force ×
.
m that has to match on the two sides. a
a s earm

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Tip: use identical one-rupee coins and a stiff plastic scale, and hang the pans with
se m
com g l a
.
thin thread so the arm length is easy to read. If the empty beam is not level to start
m a
ase
with, shift one pan slightly until it is — otherwise every reading is thrown off.

agl

co m
Pause and Ponder — Page 135
m .
m as e
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se m
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a c
m .
m a s e
e m . co agl
g l as
a

com
m .
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.co


a g l Page 29 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

7.6.3 Lever — Classes of levers

PAUSE AND PONDER

Q11 Why is it easier to open the lid of a can by using a spoon as shown in Fig. 7.35?

Spoon Lid

Rim of the can

Can

hand holding the can steady

Fig. 7.35, page 135 — redrawn sketch: the tip of a spoon is pushed under the lid of a can
and the handle is pressed, while the other hand holds the can.

The spoon works as a class-I lever with a very large mechanical advantage: the rim of the can is
the fulcrum, the edge of the lid is the load and your hand at the far end of the handle supplies
the effort.

Load arm = distance from the can's rim to the lid edge ≈ 1 cm

Effort arm = distance from the rim to your hand ≈ 10 cm

Mechanical advantage = effort arm / load arm = 10 cm / 1 cm = 10

So a hand force of 20 N produces a lifting force of

F2 = F1 × (d1/d2) = 20 N × 10 = 200 N on the lid

Page 30 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Why it happens: the lid is held by a rim that grips all round, and prising it needs a
force far larger than fingers can give. The spoon does not reduce the work, but it
multiplies the force ten times by making your hand travel about ten times as far as
the lid edge moves. F1d1 = F2d2 is satisfied throughout.

Try This: hold the spoon halfway along the handle instead of at the end. The effort
arm halves, the mechanical advantage halves, and the lid becomes noticeably
harder to lift — proof that it is the arm lengths that matter.

Q12 Why do you push an object closer to scissors fulcrum when you want to cut an
object which is hard?

Because moving the object towards the fulcrum shortens the load arm, and a shorter load arm
means a larger mechanical advantage — so the blades press on the object with a much bigger
force.

Scissors are a class-I lever: the rivet is the fulcrum.

Mechanical advantage = effort arm / load arm

Effort arm (rivet to your fingers) stays the same, say 8 cm

Object near the tip: load arm = 6 cm → MA = 8/6 = 1.3

Object near the rivet: load arm = 1 cm → MA = 8/1 = 8

With a 30 N squeeze of the fingers:

near the tip: cutting force = 30 N × 1.3 = 40 N

near the rivet: cutting force = 30 N × 8 = 240 N

Why it happens: the blades turn about the rivet, so a point close to the rivet moves
through a very small arc while your fingers move through a large one. By F1d1 =
F2d2, a small movement at the load must come with a large force. Hard materials
such as thick cardboard or a wire need that large force; soft paper does not, so it can
be cut anywhere along the blade.

Page 31 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Did you know? Wire cutters and pliers are built with permanently short jaws and
long handles for exactly this reason — their mechanical advantage is high by design.

Q13 Throughout history, many designs of perpetual machines (using wheels, weights
or magnets) have been proposed but none actually work. Why do all real machines
eventually slow down and stop? Explain in terms of work and energy.

Every real machine has friction and air resistance acting on its moving parts. These forces do
negative work continuously, draining the machine's mechanical energy into heat and sound —
forms that cannot climb back into the moving parts. With no fuel or electricity to top up the
store, the mechanical energy runs down to zero and the machine stops.

Work–energy theorem applied to the machine:

change in mechanical energy = work done by all forces

Friction and drag always oppose the motion, so their work is negative

mechanical energy after one cycle = mechanical energy before – (energy lost as heat and

sound)

Each cycle the store is smaller → motion becomes slower → finally zero

A perpetual machine would have to be one of two impossible things:

A machine that creates energy out of nothing, so that it can keep doing useful work forever.
That breaks the conservation of energy — energy can change form but never appear from
nowhere.
A machine with absolutely no friction, air resistance or sound, which merely keeps moving.
Even that would only keep going; it could do no useful work, because every joule of useful
work taken out would have to come from its own store.

Why friction cannot be undone: the chapter's Ready to Go Beyond box on page
126 makes the key point — work done against gravity, or against electric and
magnetic forces, is stored as potential energy and can be returned. Work done
against friction is not stored. It becomes disordered thermal energy spread through
the parts and the air, and no arrangement of wheels, weights or magnets can gather
it back.

Page 32 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Tip: good design can make losses small — smoother bearings, oil, streamlined
shapes — so a machine runs longer. It can never make them zero.

What If … — Page 135
7.6.3 Lever — Classes of levers

WHAT IF …

Q1 … it were possible to build a perpetual motion machine, which once started, could
continue doing useful work forever, without any fuel or electricity?

It cannot be built, and the reason is the very theorem this chapter is built on: work done =
change in energy. Doing useful work forever would mean handing out energy forever from a
store that is never refilled.

Useful work delivered in time t = W

Energy taken from the machine's store = W (nothing else supplies it)

After time t: energy left = E0 – W

As t grows, the store empties and the machine must stop.

Running forever would need E0 = infinite, or energy created from nothing.

If such a machine did exist, imagine what would follow:

No power stations, no dams, no fuel — a small box in every home would run lights, pumps
and trains forever, and there would be no fuel bills and no smoke.
But the same logic would break the rest of physics. Conservation of energy is what lets us
predict the speed of a falling ball, the reach of a rocket and the heat from a fuel. If energy
could appear from nowhere, none of those calculations would work — and yet they work
every day.

Why the idea keeps returning: a well-made wheel or magnet arrangement can
spin for a surprisingly long time, which looks like "forever" to a hopeful inventor.
Careful measurement always shows a slow, steady loss to friction, air drag and
sound. Every proposed design, when tested, has failed at exactly this point.

Page 33 of 63

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Class 9 Science Chapter 7 Work, Energy, and Simple Machines
g l AglaSem · NCERT Solutions

co m
m.
Did you know? The failure of thousands of such designs is itself scientific evidence.

m l a se
It is one of the strongest experimental supports for the law of conservation of
o
energy —.cnature has been offered every chance to break it and a g never done so.
m
has

l a se
ag
om
Revise, Reflect, Refine — Pages 136 – 138
. c ag
s e m
a
End-of-chapter exercise

REVISE, REFLECT, REFINE agl

co m
se m.
Q1 State whether True or False. (i) Work is said to be done when a force is applied, even

o m l a
if the object does not move. (ii) Lifting a bucket vertically upward results in positive
genergy is joule (J). (iv) A
m .c done on the bucket. (iii) The SI unit for both work and
a
se motionless stretched rubber band has kinetic energy. (v) Energy can change from
work
a
agl one form to another.

m a s
em
.co agl
a s
# STATEMENT
a gl TRUE / REASON
FALSE

co m
m .
e
(i) Work is done when a force is applied, False W = F × s. With s = 0 the work is zero, however hard

m l as
.co
even if the object does not move you push (Fig. 7.5, pushing a wall).

a g
se m
l a
(ii) Lifting a bucket vertically upward True Your force is upward and the displacement is upward

ag does positive work on the bucket — same direction, so W is positive.

se m
com l a
(iii) The SI unit for both work and energy True Work done appears as a change in energy, so the two
is the joule (J)
.
must share a unit. 1 J = 1 N m = 1 kg m² s⁻².
a g
a sem
A motionless stretched rubberlband
ag
(iv) False K = ½mv² and v = 0, so K = 0. It stores elastic
has kinetic energy potential energy because of its deformation.

co m
.
(v) Energy can change from one form to True Electrical → light in a bulb, chemical → mechanical

e m
as
another in muscles, mechanical → sound in a bell (Section

m l
.co a g
7.3).

se m
g l a
a Watch out for (i) and (iv): both are the traps the chapter sets. Tiredness is not work,
c
m .
and being stretched is not being in motion. The test for work is always displacement
m a s e
. co agl
along the force; the test for kinetic energy is always speed.
e m
g l as
a

co m
m .
m as e
.co


a g l Page 34 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Q2 Fill in the blanks. (i) Work done = ______ × ______ (in the direction of force). (ii) 1 joule of
work is done when a force of ______ newton displaces an object by 1 metre in the
direction of the force. (iii) The expression for kinetic energy of a body of mass m and
velocity v is ______. (iv) The potential energy of an object of mass m at a small height
h from the Earth’s surface is ______. (v) Power is defined as the ______ at which work is
done.

# ANSWER WHERE IT COMES FROM

(i) force × displacement Eq. 7.1 / 7.2: W = F × s

(ii) 1 1 J = 1 N × 1 m, page 118

(iii) ½mv² Eq. 7.6

(iv) mgh Eq. 7.8

(v) rate Eq. 7.11: P = W/t

Check the units of each formula:

W=F×s→N×m=J

K = ½mv² → kg × (m s–1)2 = kg m2 s–2 = J

U = mgh → kg × m s–2 × m = kg m2 s–2 = J

P = W/t → J / s = W (watt)

Tip: all three energy formulae must reduce to kg m² s⁻². If a formula you have
written does not, it is wrong — this is the quickest check you can run in an exam.

Q3 When a ball thrown upwards reaches its highest point, tick which of the following
statement(s) are correct? (i) The force acting on the ball is zero. (ii) The acceleration
of the ball is zero. (iii) Its kinetic energy is zero. (iv) Its potential energy is
maximum.

The correct statements are (iii) and (iv).

Page 35 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

# STATEMENT CORRECT? REASON

(i) The force acting on the ball is No Gravity never switches off. The weight mg still acts
zero downward at the top.

(ii) The acceleration of the ball is No a = F/m = mg/m = g = 10 m s⁻² downward, even at the
zero instant v = 0.

(iii) Its kinetic energy is zero Yes At the highest point v = 0, so K = ½mv² = 0.

(iv) Its potential energy is Yes h is greatest there, so U = mgh is greatest.
maximum

Mechanical energy is conserved throughout the flight:

At the throw: K = ½mu2, U = 0

At the top: K = 0, U = mghmax

Equating: mghmax = ½mu2 → hmax = u2/2g

Why (i) and (ii) trap so many students: zero velocity is confused with zero
acceleration. Velocity is the ball's speed; acceleration is how fast that speed is
changing. At the top the velocity is passing through zero on its way from up to down,
and it is changing as fast as ever — at 10 m s⁻¹ every second.

Page 36 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Q4 For each of the following situations, identify the energy transformation that takes
place: (i) a truck moving uphill, (ii) unwinding of a watch spring, (iii) photosynthesis
in green leaves, (iv) water flowing from a dam, (v) burning of a matchstick, (vi)
explosion of a fire cracker, (vii) speaking into a microphone, (viii) a glowing electric
bulb, and (ix) a solar panel.

# SITUATION ENERGY WHAT ACTUALLY HAPPENS
TRANSFORMATION

(i) A truck moving Chemical → kinetic + Diesel burns; the engine drives the wheels and
uphill gravitational potential (+ thermal) the truck also gains height, so it gains mgh.
Friction and the hot exhaust carry away the
rest.

(ii) Unwinding of a Elastic potential → mechanical The wound spring is deformed and stores
watch spring (kinetic) energy; as it unwinds it turns the gears and the
hands.

(iii) Photosynthesis in Light (solar) → chemical Sunlight absorbed by chlorophyll is stored in
green leaves the bonds of glucose — the food chain's
energy store.

(iv) Water flowing Gravitational potential → kinetic Water high in the reservoir has mgh; falling
from a dam (→ electrical in the turbine) turns it into ½mv², which spins the turbine.

(v) Burning of a Chemical → thermal + light The chemicals on the head react and release
matchstick stored bond energy as heat and a flame.

(vi) Explosion of a fire Chemical → thermal + light + A very fast reaction; the hot gases push
cracker sound + kinetic outward, so fragments and air also gain kinetic
energy.

(vii) Speaking into a Sound → electrical Air vibrations move a diaphragm, and the
microphone moving diaphragm generates a matching
electrical signal.

(viii) A glowing electric Electrical → light + thermal Current heats the filament until it glows. In a
bulb filament bulb most of the energy leaves as
heat, not light.

(ix) A solar panel Light (solar) → electrical Photocells convert sunlight directly into an
electric current.

Page 37 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

The pattern to notice: in every case the total energy is the same before and after —
only the label changes. And in almost every case some of it ends up as thermal
energy, which is the hardest form to use again.

Q5 A student is slowly lifted straight up in an elevator from the ground level to the top
floor of a building. Later, the same student climbs the staircase, all the way to the
top. Given that the height of the building is h = 72.5 m, acceleration due to gravity is
g = 10 m s–2, and student’s mass is m = 50 kg. (i) Find the gain in the potential
energy if the student is lifted straight up to the top. (ii) Find the gain in the
potential energy when the student climbs the stairs to the same top. (iii) What do
you conclude about the dependence of the potential energy on the path taken?

(i) 36 250 J. (ii) 36 250 J — exactly the same. (iii) Gravitational potential energy depends
only on the height gained, not on the path.

(i) By elevator

Formula: ΔU = mgh

ΔU = 50 kg × 10 m s–2 × 72.5 m

ΔU = 36 250 J = 3.625 × 104 J (about 36.25 kJ)

(ii) By the staircase

The starting height and the finishing height are the same, so h is still 72.5 m

ΔU = 50 kg × 10 m s–2 × 72.5 m = 36 250 J — no difference

(iii) Conclusion: the gravitational potential energy of an object depends only on its vertical
height above the chosen reference level. It does not depend on the route — straight up, up a
staircase, up a spiral ramp or up a hill road all give the same mgh.

Why it happens: gravity acts vertically downward. On the horizontal parts of a
staircase — the treads — the displacement is perpendicular to gravity, so gravity
does no work there and nothing is stored. Only the vertical parts, the risers, count,
and adding up all the risers gives exactly 72.5 m.

Page 38 of 63

as e
a
Class 9 Science Chapter 7 Work, Energy, and Simple Machines
g l AglaSem · NCERT Solutions

co m
m.
What is different: the student gets far more tired on the stairs. That is because the

m l a se
leg muscles also do work moving the body forward and back, and lose energy as
o g is still 36 250 J
.c the body. The energy stored in the Earth–student system
a
m
heat inside

l a seway.
ag
either

o m
e
. c
m of a building in a certain time. It then raises ag
s
A crane lifts a mass m to the 10th floor
la of the same building in double the time. How much
Q6

a g
the same mass to the 20th floor
more energy and power are required? Assume that the height of all floors is equal.

co m
em.

m l as
.co a g
Twice the energy is needed — that is 100 % more. The power needed is exactly the same —

se m
no extra power at all.

g l a
a
s
Let the height of one floor be h0 and the first lift take time t.
m a
m.co agl
l a se
First lift — to the 10th floor
a g
height = 10h0

co m
m .
e
E1 = mg × 10h0
m l as
.co
P1 = E1/t = 10mgh0 / t
m a g
l a se
ag
Second lift — to the 20th floor, in time 2t
se m
com g l a
height = 20h0
m . a
ase
agl
E2 = mg × 20h0 = 2E1

P2 = E2/(2t) = 20mgh0 / 2t = 10mgh0 / t = P1

co m
m .
m as e
.co
Extra energy required = E2 – E1 = 10mgh0, i.e. 100 % more
a g l
se m Extra power required = P – P = 0 W
g l a
a
2 1

.c
s e m
m a
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 39 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Why the power does not change: power is work per unit time. Doubling the height
doubles the work, but doubling the time doubles the interval over which it is spread.
The two factors of 2 cancel, so the crane's motor is working at exactly the same rate
— it is simply working for longer.

Check it yourself: put in numbers. Take m = 500 kg, h₀ = 3 m, g = 10 m s⁻², t = 30 s.
Then E₁ = 500 × 10 × 30 = 150 000 J and P₁ = 5000 W. And E₂ = 500 × 10 × 60 = 300 000
J with P₂ = 300 000 / 60 = 5000 W. Same power.

Q7 Which factors determine the energy required to raise a flag from the ground to the
top of a tall flagpole using a pulley? Does raising the flag slowly or quickly change
the amount of work done? If the speed at which the flag is raised is doubled, how
does the power requirement change? Explain your answers.

The energy depends on the mass of the flag, the height of the pole and g. Speed does not
change the work done. Doubling the speed doubles the power required.

Factors that fix the energy

E = mgh, so it depends on:

• m — the mass of the flag (and of the rope raised with it)

• h — the height of the flagpole

• g — the acceleration due to gravity

A fixed pulley has mechanical advantage 1; it only turns your downward pull into an

upward lift, so it changes neither the force nor the energy.

Slowly or quickly?

W = mgh contains no t. The work done is the same either way.

Doubling the speed

P = W/t. If the speed doubles, the time halves: t′ = t/2

P′ = W / (t/2) = 2W/t = 2P — the power requirement doubles

Page 40 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Numbers make it concrete. Take a flag of mass 0.5 kg, a pole 10 m tall, g = 10 m s–2:

HOW IT IS RAISED WORK DONE TIME POWER

Slowly, at 0.5 m s⁻¹ 0.5 × 10 × 10 = 50 J 20 s 2.5 W

Twice as fast, 1 m s⁻¹ 50 J — unchanged 10 s 5 W — doubled

Why work and power part company here: work counts how much energy is
transferred; power counts how fast. The flag ends up at the same height with the
same potential energy either way — that is the work. But the person or motor
pulling the rope must deliver it in half the time, and that is a heavier demand on the
muscles or the motor.

Q8 A man of mass 60 kg rides a scooter of mass 100 kg. He accelerates the scooter to a
velocity v. The next day, his son with a mass of 40 kg joins him as a passenger. If the
scooter reaches the same speed on both days in the same time interval, what is the
ratio of the fuel of the tank used on the two days? Assume that the energy transfer
to the scooter happens entirely due to fuel, and no other losses occur due to air
resistance and friction.

The ratio of fuel used on day 1 to day 2 is 4 : 5.

Page 41 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Day 1 — man + scooter

total mass M1 = 60 kg + 100 kg = 160 kg

starting from rest, final speed v:

E1 = ½M1v2 = ½ × 160 kg × v2 = 80v2 joule (v in m s–1)

Day 2 — man + son + scooter

total mass M2 = 60 kg + 40 kg + 100 kg = 200 kg

E2 = ½M2v2 = ½ × 200 kg × v2 = 100v2 joule

Ratio of fuel used (all the energy comes from fuel, with no other losses)

E1 : E2 = 80v2 : 100v2 = 80 : 100 = 4 : 5

Why the time interval does not enter: the fuel supplies energy, and the energy
needed is fixed by the kinetic energy gained, ½Mv². The time affects only the power
the engine must deliver, not the total fuel burnt. Because the same time is given on
both days, the powers are in the same 4 : 5 ratio too.

Tip: a very common slip is to compare 60 kg with 100 kg, or with 60 + 40 = 100 kg.
The engine has to accelerate the whole moving system, so the scooter's own 100 kg
must be counted on both days.

Q9 On a seesaw with sliding seats, a child is sitting on one side and an adult on the
other side. The adult weighs twice that of the child. The seesaw however is
balanced. Draw a figure which depicts this situation showing the distances from the
fulcrum where the child and the adult are seated.

For balance, the child must sit twice as far from the fulcrum as the adult. If the adult sits at a
distance d, the child sits at 2d.

Page 42 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Adult, weight 2W
Child, weight W

2d d
W × 2d = 2W × d — the seesaw balances

Fulcrum

A seesaw is a class-I lever. The lighter child sits twice as far from the fulcrum as the adult who weighs
twice as much.

Balance condition for a lever (Eq. 7.15):

effort × effort arm = load × load arm

Let the child's weight be W and the adult's be 2W.

W × dchild = 2W × dadult

W cancels: dchild = 2 × dadult

So if the adult sits 1 m from the fulcrum, the child must sit 2 m from it.

Check: 300 N × 2 m = 600 N m and 600 N × 1 m = 600 N m ✔

Why it works: what balances a seesaw is not weight alone but the product weight ×
distance from the fulcrum — the turning effect. Being half as heavy is made up for
exactly by sitting twice as far out.

Try This: on the seesaw of Example 7.13 (Fig. 7.34), where AC = EC = 2 m and BC = DC
= 1 m, a 15 kg child on seat A balances a 30 kg child on seat D — the same 2 : 1 rule.

Page 43 of 63

as e
a
Class 9 Science Chapter 7 Work, Energy, and Simple Machines
g l AglaSem · NCERT Solutions

co m
m.
A ball of mass 2 kg is thrown up with a velocity of 20 m s–1. (i) Identify the sign of
se
Q10

o m l a
the work done by gravity on the ball during its upward motion and its downward
g was done by air
m .c (ii) If the ball reaches a height of 19.4 m, how much awork
motion.
e
las
resistance (assume g = 10 m s–2).
ag
m

. co ag
m
(i) Negative going up, positive coming down. (ii) The air resistance did –12 J of work on the

as e
l
ball during the rise.
a g
m
(i) Sign of the work done by gravity
co
Gravity always acts downward.
se m.
o m
c displacement is upward, opposite to the force → negative
Going .up: l a
gwork (the ball slows
m a
l a se and loses kinetic energy)
a g down

s
Coming down: displacement is downward, along the force → positive work (the ball speeds
m a
up and gains kinetic energy)
m.co agl
l a se
a g
(ii) Work done by air resistance during the rise
com
m = 2 kg, u = 20 m s–1, h = 19.4 m, g = 10 m s–2
m .
m as e
.co a g l
s m energy at the throw:
eKinetic
gl a
a
Ki = ½mu2 = ½ × 2 kg × (20 m s–1)2 = 400 J
se m
m stops for an instant)
coball g l a
. a
em
Kinetic energy at the top: Kf = 0 J (the

a s
agl
Work done by gravity over the rise:

co m
Wgravity = –mgh = –(2 kg × 10 m s–2 × 19.4 m) = –388 J
m .
m as e
.co a g l
se m Work–energy theorem for the whole rise:
g l a
a c
m .
e
Wgravity + Wair = Kf – Ki
m a s
(–388 J) + Wair = 0 J – 400 J = –400 J
e m . co agl
g l as
a
Wair = –400 J + 388 J = –12 J

co m
m .
m ase
.co


a g l Page 44 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Why the answer is negative, and why it is small: air resistance opposes the
motion at every instant, so its work must be negative — it is a drain, never a source.
Those 12 J are exactly the energy that has left the ball as heat and swirling air. Notice
the consequence: without air the ball would have risen to u²/2g = 400/20 = 20 m. The
12 J lost is what cost it the last 0.6 m.

Check it yourself: 12 J ÷ (mg) = 12 J ÷ 20 N = 0.6 m — the shortfall in height. The
energy bookkeeping closes exactly.

Q11 A 10.0 kg block is moving on horizontal floor with negligible friction. As shown in
the Fig. 7.37, a variable force is applied on the block in its direction of motion from
its position at 0 m till 4 m. If the block had a kinetic energy of 180 J when it was at
0 m, find the block’s speed (i) at 0 m, and (ii) at 4 m. Does the block have negative
acceleration in any portion of its motion?

50
Force (N)

0 1 2 3 4
Displacement (m)

Fig. 7.37, page 138: the force applied on the block against its displacement.

(i) 6 m s⁻¹ at 0 m. (ii) About 8.12 m s⁻¹ at 4 m. No — the acceleration is never negative,
though it does decrease between 3 m and 4 m.

Page 45 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

50
Force (N)

shaded area = work = 150 J

0 1 2 3 4
Displacement (m)

The work done by the varying force is the area of the trapezium under the force–displacement graph
of Fig. 7.37.

(i) Speed at 0 m

K = ½mv2 → v = √(2K/m)

v = √(2 × 180 J / 10.0 kg) = √(36 m2 s–2) = 6 m s–1

Work done by the force from 0 m to 4 m = area under the graph

The shape is a trapezium: parallel sides 4 m (at the bottom) and 2 m (the flat top, from 1 m to

3 m), height 50 N.

W = ½ × (4 m + 2 m) × 50 N = ½ × 6 m × 50 N = 150 J

Check by pieces: ½(1 m)(50 N) + (2 m)(50 N) + ½(1 m)(50 N) = 25 + 100 + 25 = 150 J ✔

(ii) Speed at 4 m — work–energy theorem

K at 4 m = K at 0 m + work done = 180 J + 150 J = 330 J

v = √(2 × 330 J / 10.0 kg) = √(66 m2 s–2) = 8.12 m s–1

Page 46 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Is the acceleration ever negative? No.

a = F/m, and the graph shows F is positive (along the motion) at every point from 0 m to 4

m.

0 to 1 m: F rises 0 → 50 N, so a rises 0 → 5 m s–2

1 to 3 m: F = 50 N, so a = 50 N / 10.0 kg = 5 m s–2, constant

3 to 4 m: F falls 50 N → 0, so a falls 5 m s–2 → 0 — smaller, but still positive

The distinction that matters: a decreasing force is not a backward force. Between
3 m and 4 m the block is still being pushed forward, so it keeps gaining speed — just
less rapidly. The acceleration would only turn negative if the graph dipped below the
axis, and it never does.

Q12 The gravitational attraction on the surface of the Moon (lunar surface) is about
1/6th of that on the surface of the Earth. An astronaut can throw a ball up to a
height of 8 m from the surface of the Earth. How far up will the ball thrown with
the same upward velocity travel from the surface of the Moon?

The ball will rise to 48 m on the Moon — six times as high.

Page 47 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

The astronaut throws with the same speed u both times, so the ball starts with the same

kinetic energy.

At the highest point all of it has become potential energy:

½mu2 = mgh → h = u2 / 2g

On the Earth

hE = u2 / (2gE) = 8 m

On the Moon, gM = gE/6

hM = u2 / (2gM) = u2 / (2 × gE/6) = 6 × u2/(2gE) = 6 hE

hM = 6 × 8 m = 48 m

With numbers, taking gE = 10 m s–2:

u = √(2gEhE) = √(2 × 10 × 8) = √160 ≈ 12.6 m s–1

gM = 10/6 ≈ 1.67 m s–2

hM = u2/(2gM) = 160 / (2 × 1.67) = 48 m ✔

Why it happens: the same throw gives the ball the same kinetic energy ½mu². On
the Moon each metre of rise costs only mgMh = one-sixth as much potential energy,
so the same energy buys six times the height. Notice that the mass of the ball never
enters — it cancels on both sides.

Did you know? This is why astronauts on the Moon bounced along in long, slow
hops. The same push of the legs lifted them six times higher.

Page 48 of 63

as e
Class 9 Science Chapter 7 Work, Energy, and Simple Machines
a g l AglaSem · NCERT Solutions

co m
m.
A 1000 kg car is moving along a road at a constant speed. Suddenly, the driver
se
Q13

o m l a
notices some obstruction ahead and applies the brakes to come to a complete
g from the instant
stop..cThe graphical representation of motion of the car starting
m a
e driver spots the traffic ahead is shown in Fig. 7.38. (i) Describe how the car
las
the
ag moves between positions A and B. (ii) Calculate the kinetic energy of the car at A.
(iii) State the work done by the brakes in bringing the car to a halt between B and

co m
. ag
C. (iv) What does the kinetic energy of the car transform into?

e m
g l as
a

co m
e m.
m A B
l as
.co 35 a g
a sem
agl
Speed (m s⁻¹)

m a s
m .co agl
l a se
a g
C
c3om
m .
0 1 2
as e
. com a g l
m
Time (s)
a s e
agl Fig. 7.38, page 138: speed–time graph of the car from the instant the driver spots the
obstruction (A) to the halt (C).
se m
com g l a
m . a
ase
ANSWER agl

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 49 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

A B
35
Speed (m s⁻¹)

area = distance travelled

C

0 1 2 3
Time (s)

A to B: constant speed for one second — the driver's reaction time. B to C: uniform braking to rest in
two seconds.

(i) Between A and B the car moves in a straight line at a constant speed of 35 m s–1.

The graph is horizontal from t = 0 s to t = 1 s, so the speed does not change.

Acceleration = 0, so the net force on the car is zero — the brakes have not been applied yet.

This one second is the driver's reaction time, between spotting the obstruction and pressing

the pedal.

Distance covered = 35 m s–1 × 1 s = 35 m travelled before braking even begins.

(ii) Kinetic energy at A

K = ½mv2

K = ½ × 1000 kg × (35 m s–1)2

K = 500 kg × 1225 m2 s–2

K = 612 500 J = 6.125 × 105 J

Page 50 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

(iii) Work done by the brakes from B to C

At B the car still has 612 500 J (the speed is unchanged at 35 m s–1).

At C the car is at rest: K = 0 J.
Work–energy theorem: work done = change in kinetic energy

W = 0 J – 612 500 J = –612 500 J

The negative sign shows the braking force opposes the motion.

Cross-check: braking distance = area of the triangle = ½ × 2 s × 35 m s–1 = 35 m
braking force = 612 500 J ÷ 35 m = 17 500 N

and from Newton's second law: a = (0 – 35)/2 = –17.5 m s–2, F = 1000 × 17.5 = 17 500 N ✔

(iv) The kinetic energy is transformed mainly into thermal energy — heat in the brake pads
and discs, in the tyres and in the road surface — together with a small amount of sound energy
(the screech of the tyres).

Why this matters on the road: the total stopping distance is 35 m of reaction plus
35 m of braking — 70 m in all. And because K goes as v², a car at 70 m s⁻¹ would
need four times the braking energy to be removed, so about four times the braking
distance.

Page 51 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Q14 The potential energy-displacement graph of a 0.5 kg ball moving along a
frictionless track is shown in Fig. 7.39. At O, the velocity of the ball is 0 m s–1 and
potential energy is 30 J. Calculate the velocity of the ball at P, Q and R.

40 R
Potential Energy (J)

O
30
Q

20
P

10

Displacement (m)

Fig. 7.39, page 138: potential energy of the 0.5 kg ball against its displacement along
the frictionless track.

At P the ball moves at about 6.3 m s⁻¹, at Q it is momentarily at rest (0 m s⁻¹), and it can
never reach R at all.

Page 52 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

R lies above the line —
unreachable
R
40
Potential Energy (J)

O total mechanical energy
Q = 30 J
30

P
20

10

Displacement (m)

The dashed line is the ball's fixed store of 30 J. It can only visit points where the curve lies below that
line.

Step 1 — find the total mechanical energy

At O: v = 0 m s–1, so K = 0 J; and U = 30 J
Total mechanical energy E = K + U = 0 + 30 = 30 J

The track is frictionless, so E stays 30 J everywhere.

Step 2 — at each point, K = E – U, then v = √(2K/m), with m = 0.5 kg

POINT U FROM THE GRAPH K = 30 J – U V = √(2K/M)

P 20 J 10 J √(2 × 10 / 0.5) = √40 ≈ 6.3 m s⁻ ¹

Q 30 J 0J 0 m s⁻ ¹ — momentarily at rest

R 40 J –10 J impossible — the ball never gets to R

Page 53 of 63

as e
a
Class 9 Science Chapter 7 Work, Energy, and Simple Machines
g l AglaSem · NCERT Solutions

co m
e m.
At P: K = 30 J – 20 J = 10 J
m l as
.co
v = √(2K/m) = √(2 × 10 J / 0.5 kg) = √(40 m2 s–2) = 6.32 m s–1
m a g
l a se
g
aAt Q: K = 30 J – 30 J = 0 J

com
v = √(0) = 0 m s–1
e m . ag
g l as
At R: K = 30 J – 40 J = –10 J
a

co m
m.
But K = ½mv2 can never be negative, so R cannot be reached.

m as e
.co a g l
a s em actually happens: the ball rolls away from O, speeds up through the dips,
gl slows as it climbs and arrives at Q with exactly zero speed. Q is its turning point. It
What
a
s
then rolls back the way it came, passing P again at 6.3 m s⁻¹ and returning to O with
m a
.co agl
zero speed once more. Because there is no friction, it repeats this journey forever

se m
a
between O and Q.

a g l

m
Tip: on any potential-energy graph, draw the horizontal line at the total energy. The

. co
m
object can only move in the regions where the curve lies below that line; the points

m as e
.co l
where the curve meets the line are its turning points.
a g
sem
g l a
a
Q15 A coconut of mass 1.5 kg falls from the top of a coconut tree onto the wet sand on
se m
com
a beach. The height of the tree is 10 m. On impact, the coconut comes to rest by
g l a
m . a
e
s Assume that the average resistive force of sand is 3000
making a depression in the sand. (i) Calculate the velocity of the coconut just
a(ii)
a g
before it hits the sand.l
N and all of the coconut’s energy is used to create the depression in the sand.

m
Calculate the depth of the depression the coconut makes in the sand. Assume g =

. co
m
10 m s–2.

as e
. com a g l
sem

a
agl
(i) About 14.1 m s⁻¹. (ii) The depression is about 0.05 m, that is 5 cm deep.

.c
s e m
m a
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 54 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

(i) Speed just before impact

m = 1.5 kg, h = 10 m, g = 10 m s–2

The coconut starts from rest, and only gravity acts, so mechanical energy is conserved:

potential energy at the top = kinetic energy at the sand

mgh = ½mv2

m cancels: v2 = 2gh = 2 × 10 m s–2 × 10 m = 200 m2 s–2

v = √200 = 14.14 m s–1 ≈ 14.1 m s–1

(ii) Depth of the depression

Energy the coconut brings to the sand:

E = mgh = 1.5 kg × 10 m s–2 × 10 m = 150 J

(check: ½mv2 = ½ × 1.5 kg × 200 m2 s–2 = 150 J ✔)

The sand pushes back with F = 3000 N through the depth d, doing negative work on the

coconut:

work done by the sand = –F × d

The coconut is brought to rest, so this work must remove all 150 J:

F × d = 150 J

3000 N × d = 150 J

d = 150 J / 3000 N = 150 N m / 3000 N

d = 0.05 m = 5 cm

Why the sand saves the coconut: the same 150 J is removed in either case, but the
force depends on the stopping distance. On sand, F = 150 J ÷ 0.05 m = 3000 N. On a
concrete floor the coconut would stop in perhaps 1 mm, needing F = 150 J ÷ 0.001 m
= 150 000 N — fifty times greater, and enough to smash it. This is exactly why we use
crash mats, airbags and packing material.

Page 55 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

A finer point: while the coconut sinks the extra 0.05 m it falls a little further, gaining
mgd = 1.5 × 10 × 0.05 = 0.75 J more. Including it gives 3000d = 150 + 15d, so d =
150/2985 = 0.0503 m — still 5 cm to two significant figures, exactly as the question
intends.

The Journey Beyond — Page 139
Project work

THE JOURNEY BEYOND

Q1 Remove both the ends from a pen so that the refill can slide freely through the
barrel (Fig. 7.40). Fix the pen cap to the side of the barrel and attach a rubber band
to the clip of the cap. Connect the free end of the rubber band to the refill using a
safety pin. Stretch and release the rubber band. The refill shoots out, showing the
conversion of elastic potential energy into kinetic energy. Repeat with different
amounts of stretch, and observe how the distance travelled changes. Is there a
relationship between the stretch and the distance travelled?

Refill
Pen cap fixed to the side

Rubber band

Safety pin

Pen barrel (both ends removed)

Fig. 7.40, page 139 — redrawn sketch of the pen launcher: the cap is fixed to the side of
the barrel and a rubber band runs from the cap’s clip to a safety pin attached to the
refill, which slides freely inside the barrel.

Yes. The farther you stretch the band, the farther the refill flies — and the distance grows
much faster than the stretch does, roughly as the square of the extension.

Page 56 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

How to do the experiment properly

1. Mark the barrel in centimetres so you can pull the refill back by a measured amount x each
time.
2. Fire the refill horizontally from the same height and from the same launch spot every time —
tape the pen to the edge of a table.
3. Measure the horizontal distance R from the table edge to where the refill first lands.
4. Take three shots at each stretch and use the average, to reduce the error.

STRETCH X (CM) DISTANCE R (CM) — SAMPLE READINGS R÷X R ÷ X²

2 40 20 10

3 90 30 10

4 160 40 10

Why R goes as x2

Elastic potential energy stored in the stretched band grows as the square of the extension: U

∝ x2

All of it becomes kinetic energy of the refill: ½mv2 = U ∝ x2

so v2 ∝ x2, i.e. v ∝ x

For a horizontal launch from a fixed height, the time of flight t is the same every shot,

so for a horizontal launch off a table the time of flight t is fixed and R = v × t ∝ v ∝ x

but for a launch at an angle the projectile formula gives R ∝ v2 ∝ x2

What to expect in your own data: if you fire horizontally off a table, R rises in
proportion to x. If you fire at an angle across the floor, R rises roughly as x². Either
way, doubling the stretch more than doubles the reach — and a graph of R against
x² (or against x) that comes out as a straight line is the proof.

Safety: never aim the refill at anyone's face, and wear spectacles if you have them.
Fire along the floor or into a cardboard box.

Page 57 of 63

Class 9 Science Chapter 7 Work, Energy, and Simple Machines AglaSem · NCERT Solutions

Q2 Construct one or more simple machines, or a combination of them (lever, pulley,
and inclined plane) using easily available materials, such as cardboard, wooden
strips or rulers, pencils or bolts (to act as a fulcrum), thread or rope, small pulleys
(or two bottle caps stuck together), and paper cups to hold small weights. Be
imaginative in your design. Use your model to lift or move a small load, measure
the effort and the load, and calculate the mechanical advantage.

Build any one of these three, then measure the effort with a spring balance (or by counting
identical coins) and use MA = load ÷ effort.

MACHINE HOW TO BUILD IT WHAT TO EXPECTED
MEASURE MECHANICAL
ADVANTAGE

Lever A 30 cm ruler resting across a pencil. Load arm d₂, effort MA = d₁/d₂; with d₁ = 24 cm
Load (a stone in a paper cup) near the arm d₁, and the effort and d₂ = 6 cm, MA = 4
pencil, effort at the far end. needed

Inclined A stiff cardboard ramp of length L from L, h, and the spring MA = L/h; with L = 60 cm
plane the floor to a stack of books of height h. balance reading and h = 15 cm, MA = 4
Pull a toy car up with a spring balance.

Pulley Two bottle caps glued back to back on a Weight of load, force MA ≈ 1 for a single fixed
nail axle, hung from a hook. Thread over needed on the free pulley — it changes only the
the groove, load on one side, effort on end direction
the other.

Sample answer — the ruler lever

Load = a 200 g stone, so load = mg = 0.2 kg × 10 m s–2 = 2 N

Load arm d2 = 6 cm, effort arm d1 = 24 cm

Predicted effort: F1 = F2 × d2/d1 = 2 N × 6/24 = 0.5 N

Measured effort (spring balance) = 0.6 N

Mechanical advantage = load ÷ effort = 2 N ÷ 0.6 N = 3.3

Ideal MA = d1/d2 = 24/6 = 4

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Class 9 Science Chapter 7 Work, Energy, and Simple Machines
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