NCERT Solutions for Class 9th Science Chapter 7 End-of-chapter exercise — Revise, Reflect, Refine

Book page 136 – 138 Updated on2026-09-08

Q1.

State whether True or False. (i) Work is said to be done when a force is applied, even if the object does not move. (ii) Lifting a bucket vertically upward results in positive work done on the bucket. (iii) The SI unit for both work and energy is joule (J). (iv) A motionless stretched rubber band has kinetic energy. (v) Energy can change from one form to another.

Answer

#	Statement	True / False	Reason
(i)	Work is done when a force is applied, even if the object does not move	False	W = F × s. With s = 0 the work is zero, however hard you push (Fig. 7.5, pushing a wall).
(ii)	Lifting a bucket vertically upward does positive work on the bucket	True	Your force is upward and the displacement is upward — same direction, so W is positive.
(iii)	The SI unit for both work and energy is the joule (J)	True	Work done appears as a change in energy, so the two must share a unit. 1 J = 1 N m = 1 kg m² s⁻².
(iv)	A motionless stretched rubber band has kinetic energy	False	K = ½mv² and v = 0, so K = 0. It stores elastic potential energy because of its deformation.
(v)	Energy can change from one form to another	True	Electrical → light in a bulb, chemical → mechanical in muscles, mechanical → sound in a bell (Section 7.3).

Watch out for (i) and (iv): both are the traps the chapter sets. Tiredness is not work, and being stretched is not being in motion. The test for work is always displacement along the force; the test for kinetic energy is always speed.

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

#	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.

Answer

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

#	Statement	Correct?	Reason
(i)	The force acting on the ball is zero	No	Gravity never switches off. The weight mg still acts downward at the top.
(ii)	The acceleration of the ball is zero	No	a = F/m = mg/m = g = 10 m s⁻² downward, even at the 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 maximum	Yes	h is greatest there, so U = mgh is greatest.

Mechanical energy is conserved throughout the flight:
At the throw: K = ½mu², U = 0
At the top: K = 0, U = mgh_max
Equating: mgh_max = ½mu² → h_max = 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.

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.

Answer

#	Situation	Energy transformation	What actually happens
(i)	A truck moving uphill	Chemical → kinetic + gravitational potential (+ thermal)	Diesel burns; the engine drives the wheels and the truck also gains height, so it gains mgh. Friction and the hot exhaust carry away the rest.
(ii)	Unwinding of a watch spring	Elastic potential → mechanical (kinetic)	The wound spring is deformed and stores energy; as it unwinds it turns the gears and the hands.
(iii)	Photosynthesis in green leaves	Light (solar) → chemical	Sunlight absorbed by chlorophyll is stored in the bonds of glucose — the food chain's energy store.
(iv)	Water flowing from a dam	Gravitational potential → kinetic (→ electrical in the turbine)	Water high in the reservoir has mgh; falling turns it into ½mv², which spins the turbine.
(v)	Burning of a matchstick	Chemical → thermal + light	The chemicals on the head react and release stored bond energy as heat and a flame.
(vi)	Explosion of a fire cracker	Chemical → thermal + light + sound + kinetic	A very fast reaction; the hot gases push outward, so fragments and air also gain kinetic energy.
(vii)	Speaking into a microphone	Sound → electrical	Air vibrations move a diaphragm, and the moving diaphragm generates a matching electrical signal.
(viii)	A glowing electric bulb	Electrical → light + thermal	Current heats the filament until it glows. In a 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.

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?

Answer

(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.

What is different: the student gets far more tired on the stairs. That is because the leg muscles also do work moving the body forward and back, and lose energy as heat inside the body. The energy stored in the Earth–student system is still 36 250 J either way.

Q6.

A crane lifts a mass m to the 10th floor of a building in a certain time. It then raises the same mass to the 20th floor of the same building in double the time. How much more energy and power are required? Assume that the height of all floors is equal.

Answer

Twice the energy is needed — that is 100 % more. The power needed is exactly the same — no extra power at all.

Let the height of one floor be h₀ and the first lift take time t.

First lift — to the 10th floor
height = 10h₀
E₁ = mg × 10h₀
P₁ = E₁/t = 10mgh₀ / t

Second lift — to the 20th floor, in time 2t
height = 20h₀
E₂ = mg × 20h₀ = 2E₁
P₂ = E₂/(2t) = 20mgh₀ / 2t = 10mgh₀ / t = P₁

Extra energy required = E₂ – E₁ = 10mgh₀, i.e. 100 % more
Extra power required = P₂ – P₁ = 0 W

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.

Answer

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

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.

Answer

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

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.

Answer

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.

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 × d_child = 2W × d_adult
W cancels: d_child = 2 × d_adult

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.

Q10.

A ball of mass 2 kg is thrown up with a velocity of 20 m s–1. (i) Identify the sign of the work done by gravity on the ball during its upward motion and its downward motion. (ii) If the ball reaches a height of 19.4 m, how much work was done by air resistance (assume g = 10 m s–2).

Answer

(i) Negative going up, positive coming down. (ii) The air resistance did –12 J of work on the ball during the rise.

(i) Sign of the work done by gravity
Gravity always acts downward.
Going up: displacement is upward, opposite to the force → negative work (the ball slows down and loses kinetic energy)
Coming down: displacement is downward, along the force → positive work (the ball speeds up and gains kinetic energy)

(ii) Work done by air resistance during the rise
m = 2 kg, u = 20 m s^–1, h = 19.4 m, g = 10 m s^–2

Kinetic energy at the throw:
K_i = ½mu² = ½ × 2 kg × (20 m s^–1)² = 400 J
Kinetic energy at the top: K_f = 0 J (the ball stops for an instant)

Work done by gravity over the rise:
W_gravity = –mgh = –(2 kg × 10 m s^–2 × 19.4 m) = –388 J

Work–energy theorem for the whole rise:
W_gravity + W_air = K_f – K_i
(–388 J) + W_air = 0 J – 400 J = –400 J
W_air = –400 J + 388 J = –12 J

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?

Answer

(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.

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

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?

Answer

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

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
h_E = u² / (2g_E) = 8 m

On the Moon, g_M = g_E/6
h_M = u² / (2g_M) = u² / (2 × g_E/6) = 6 × u²/(2g_E) = 6 h_E
h_M = 6 × 8 m = 48 m

With numbers, taking g_E = 10 m s^–2:

u = √(2g_Eh_E) = √(2 × 10 × 8) = √160 ≈ 12.6 m s^–1
g_M = 10/6 ≈ 1.67 m s^–2
h_M = u²/(2g_M) = 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 mg_Mh = 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.

Q13.

A 1000 kg car is moving along a road at a constant speed. Suddenly, the driver notices some obstruction ahead and applies the brakes to come to a complete stop. The graphical representation of motion of the car starting from the instant the driver spots the traffic ahead is shown in Fig. 7.38. (i) Describe how the car 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 C. (iv) What does the kinetic energy of the car transform into?

Answer

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

(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.

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.

Answer

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.

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	0 J	0 m s⁻¹ — momentarily at rest
R	40 J	–10 J	impossible — the ball never gets to R

At P: K = 30 J – 20 J = 10 J
v = √(2K/m) = √(2 × 10 J / 0.5 kg) = √(40 m² s^–2) = 6.32 m s^–1

At Q: K = 30 J – 30 J = 0 J
v = √(0) = 0 m s^–1

At R: K = 30 J – 40 J = –10 J
But K = ½mv² can never be negative, so R cannot be reached.

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

Tip: on any potential-energy graph, draw the horizontal line at the total energy. The object can only move in the regions where the curve lies below that line; the points where the curve meets the line are its turning points.

Q15.

A coconut of mass 1.5 kg falls from the top of a coconut tree onto the wet sand on a beach. The height of the tree is 10 m. On impact, the coconut comes to rest by making a depression in the sand. (i) Calculate the velocity of the coconut just before it hits the sand. (ii) Assume that the average resistive force of sand is 3000 N and all of the coconut’s energy is used to create the depression in the sand. Calculate the depth of the depression the coconut makes in the sand. Assume g = 10 m s–2.

Answer

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

(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.

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.

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