Class 9 Science NCERT Solution chapter 7 Work, Energy, and Simple Machines

Revise, Reflect, Refine — Complete Solutions

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🟦 QUESTION 1

State whether the following statements are 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

(i) False

Work is done only when the applied force produces displacement. If the object does not move, the work done is zero.$$W=F\times s$$W=F×s

When $s=0$s=0,$$W=0$$W=0

(ii) True

While lifting a bucket, the applied force and displacement are both upward. Therefore, positive work is done on the bucket.

(iii) True

The SI unit of both work and energy is the joule (J).

(iv) False

A motionless stretched rubber band has elastic potential energy, not kinetic energy.

(v) True

Energy can be transformed from one form into another, such as electrical energy into light and heat energy in a bulb.

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🟦 QUESTION 2

Fill in the blanks.

(i) Work done = ______ × ______ (in the direction of force).
(ii) One 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 $m$m and velocity $v$v is ______.
(iv) The potential energy of an object of mass $m$m at a small height $h$h from the Earth’s surface is ______.
(v) Power is defined as the ______ at which work is done.

🟩 ANSWER

(i) Work done = Force × Displacement$$W=F\times s$$W=F×s

(ii) One newton

(iii)$$K=\frac{1}{2}mv^2$$K=21​mv2

(iv)$$U=mgh$$U=mgh

(v) Power is the rate at which work is done.$$P=\frac{W}{t}$$P=tW​

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🟦 QUESTION 3

When a ball thrown upwards reaches its highest point, tick the correct statement(s):

(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

Correct statements: (iii) and (iv)

At the highest point:

• The instantaneous velocity of the ball is zero.
• Therefore, its kinetic energy is zero.

$$K=\frac{1}{2}mv^2=0$$K=21​mv2=0

• Its height is maximum, so its gravitational potential energy is maximum.

$$U=mgh$$U=mgh

However, the gravitational force and acceleration due to gravity still act downward. Therefore, statements (i) and (ii) are incorrect.

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🟦 QUESTION 4

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 firecracker
(vii) Speaking into a microphone
(viii) A glowing electric bulb
(ix) A solar panel

🟩 ANSWER

Situation	Energy transformation
(i) Truck moving uphill	Chemical energy of fuel → Mechanical energy and gravitational potential energy
(ii) Unwinding of a watch spring	Elastic potential energy → Kinetic energy
(iii) Photosynthesis	Light energy → Chemical energy
(iv) Water flowing from a dam	Gravitational potential energy → Kinetic energy
(v) Burning matchstick	Chemical energy → Heat and light energy
(vi) Firecracker explosion	Chemical energy → Heat, light, sound and kinetic energy
(vii) Speaking into a microphone	Sound energy → Electrical energy
(viii) Glowing electric bulb	Electrical energy → Light and heat energy
(ix) Solar panel	Solar or light energy → Electrical energy

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🟦 QUESTION 5

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 mh=72.5\text{ m}h=72.5 m, acceleration due to gravity is g=10 m s−2g=10\text{ m s}^{-2}g=10 m s−2, and the student’s mass is m=50 kgm=50\text{ kg}m=50 kg:

(i) Find the gain in potential energy if the student is lifted straight up to the top.
(ii) Find the gain in potential energy when the student climbs the stairs to the same top.
(iii) What do you conclude about the dependence of potential energy on the path taken?

🟩 ANSWER

Given:$$m=50\text{ kg},\quad g=10\text{ m s}^{-2},\quad h=72.5\text{ m}$$m=50 kg,g=10 m s−2,h=72.5 m

Potential energy gained:$$U=mgh$$U=mgh$$U=50\times10\times72.5$$U=50×10×72.5$$U=36,250\text{ J}$$U=36,250 J

(i) By elevator$$\boxed{U=36,250\text{ J}}$$U=36,250 J​

(ii) By staircase

The initial and final heights are the same.$$\boxed{U=36,250\text{ J}}$$U=36,250 J​

(iii) Conclusion

The gain in gravitational potential energy does not depend on the path followed. It depends only on:

• Mass of the object
• Acceleration due to gravity
• Vertical height gained

Therefore, the student gains the same potential energy in both cases.

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🟦 QUESTION 6

A crane lifts a mass mmm 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. Assume that the height of all floors is equal. How much more energy and power are required?

🟩 ANSWER

Let the height of the 10th floor be $h$h and the time taken be $t$t.

Energy required to reach the 10th floor:$$E_1=mgh$$E1​=mgh

The height of the 20th floor is $2h$2h.

Energy required to reach the 20th floor:$$E_2=mg(2h)$$E2​=mg(2h)$$E_2=2mgh=2E_1$$E2​=2mgh=2E1​

Therefore, the energy required is twice as much.

Power to reach the 10th floor:$$P_1=\frac{E_1}{t}$$P1​=tE1​​

Time taken to reach the 20th floor is $2t$2t.$$P_2=\frac{E_2}{2t}$$P2​=2tE2​​$$P_2=\frac{2E_1}{2t}$$P2​=2t2E1​​$$P_2=\frac{E_1}{t}=P_1$$P2​=tE1​​=P1​

Final Answer:

• Energy required becomes twice.
• Power required remains the same.

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🟦 QUESTION 7

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.

🟩 ANSWER

The energy required to raise the flag depends on:

• Mass of the flag, $m$m
• Acceleration due to gravity, $g$g
• Vertical height of the flagpole, $h$h

$$W=mgh$$W=mgh

Raising the flag slowly or quickly does not change the work done because its mass and vertical height remain the same.

Power is:$$P=\frac{W}{t}$$P=tW​

If the speed is doubled, the time taken becomes half.$$P’=\frac{W}{t/2}=2\frac{W}{t}$$P′=t/2W​=2tW​$$P’=2P$$P′=2P

Therefore, doubling the speed doubles the power requirement, while the work done remains unchanged.

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🟦 QUESTION 8

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

🟩 ANSWER

First day:

Total mass:$$m_1=100+60=160\text{ kg}$$m1​=100+60=160 kg

Energy transferred:$$K_1=\frac{1}{2}(160)v^2$$K1​=21​(160)v2

Second day:

Total mass:$$m_2=100+60+40=200\text{ kg}$$m2​=100+60+40=200 kg

Energy transferred:$$K_2=\frac{1}{2}(200)v^2$$K2​=21​(200)v2

Ratio:$$K_1:K_2 = 160:200$$K1​:K2​=160:200$$K_1:K_2=4:5$$K1​:K2​=4:5

Since there are no energy losses, fuel used is proportional to the energy transferred.

Final Answer:$$\boxed{\text{Fuel used on first day : second day}=4:5}$$Fuel used on first day : second day=4:5​$$\boxed{\text{Energy transferred on first day : second day}=4:5}$$Energy transferred on first day : second day=4:5​

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🟦 QUESTION 9

On a seesaw with sliding seats, a child is sitting on one side and an adult on the other side. The adult weighs twice as much as the child. The seesaw is balanced. Draw a figure showing the distances from the fulcrum at which the child and adult are seated.

🟩 ANSWER

For a balanced seesaw:$$\text{Effort}\times\text{Effort arm} = \text{Load}\times\text{Load arm}$$Effort×Effort arm=Load×Load arm

Let the child’s mass be $m$m. Then the adult’s mass is $2m$2m.

If the adult sits at a distance $d$d from the fulcrum:$$m\times d_c=2m\times d$$m×dc​=2m×d$$d_c=2d$$dc​=2d

Therefore, the child must sit twice as far from the fulcrum as the adult.

Diagram:

Child of mass m                   Adult of mass 2m
        ↓                                ↓
────────●──────────────▲────────●────────
       2d            Fulcrum     d

The child sits at distance $2d$2d, while the adult sits at distance $d$d on the opposite side.

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🟦 QUESTION 10

A ball of mass 2 kg is thrown upwards with a velocity of 20 m s−120\text{ m s}^{-1}20 m s−1.

(i) Identify the sign of the work done by gravity on the ball during its upward motion and downward motion.
(ii) If the ball reaches a height of 19.4 m, how much work was done by air resistance? Assume $g=10\text{ m s}^{-2}$g=10 m s−2.

🟩 ANSWER

(i) Sign of work done by gravity

During upward motion:

• Displacement is upward.
• Gravitational force is downward.

Therefore, work done by gravity is negative.

During downward motion:

• Displacement and gravitational force are both downward.

Therefore, work done by gravity is positive.

(ii) Work done by air resistance

Initial kinetic energy:$$K_i=\frac{1}{2}mv^2$$Ki​=21​mv2$$K_i=\frac{1}{2}\times2\times20^2$$Ki​=21​×2×202$$K_i=400\text{ J}$$Ki​=400 J

At the highest point, final kinetic energy is zero.

Potential energy gained:$$U=mgh$$U=mgh$$U=2\times10\times19.4$$U=2×10×19.4$$U=388\text{ J}$$U=388 J

Initial mechanical energy:$$E_i=400\text{ J}$$Ei​=400 J

Final mechanical energy:$$E_f=388\text{ J}$$Ef​=388 J

Work done by air resistance:$$W_{\text{air}}=E_f-E_i$$Wair​=Ef​−Ei​$$W_{\text{air}}=388-400$$Wair​=388−400$$W_{\text{air}}=-12\text{ J}$$Wair​=−12 J

Final Answer:$$\boxed{W_{\text{air}}=-12\text{ J}}$$Wair​=−12 J​

The negative sign shows that air resistance acts opposite to the motion.

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🟦 QUESTION 11

A 10.0 kg block is moving on a horizontal floor with negligible friction. As shown in Fig. 7.37, a variable force is applied on the block in its direction of motion from its position at 0 m to 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
(ii) At 4 m

Does the block have negative acceleration in any portion of its motion?

🟩 ANSWER

(i) Speed at 0 m

$$K=\frac{1}{2}mv^2$$K=21​mv2$$180=\frac{1}{2}\times10\times v^2$$180=21​×10×v2$$180=5v^2$$180=5v2$$v^2=36$$v2=36$$\boxed{v=6\text{ m s}^{-1}}$$v=6 m s−1​

Work done from 0 m to 4 m

Work done equals the area under the force-displacement graph in Fig. 7.37.

From 0 m to 1 m: triangle$$W_1=\frac{1}{2}\times1\times50=25\text{ J}$$W1​=21​×1×50=25 J

From 1 m to 3 m: rectangle$$W_2=2\times50=100\text{ J}$$W2​=2×50=100 J

From 3 m to 4 m: triangle$$W_3=\frac{1}{2}\times1\times50=25\text{ J}$$W3​=21​×1×50=25 J

Total work:$$W=25+100+25$$W=25+100+25$$W=150\text{ J}$$W=150 J

Using the work-energy theorem:$$K_f-K_i=W$$Kf​−Ki​=W$$K_f=180+150$$Kf​=180+150$$K_f=330\text{ J}$$Kf​=330 J

(ii) Speed at 4 m

$$330=\frac{1}{2}\times10\times v^2$$330=21​×10×v2$$330=5v^2$$330=5v2$$v^2=66$$v2=66$$v=\sqrt{66}$$v=66​$$\boxed{v\approx8.12\text{ m s}^{-1}}$$v≈8.12 m s−1​

Does the block have negative acceleration?

No. The applied force remains in the direction of motion throughout the journey.

From 3 m to 4 m, the force decreases, but it remains positive. Therefore, the acceleration decreases but does not become negative.

Final Answer:

• Speed at 0 m = 6 m s⁻¹
• Speed at 4 m ≈ 8.12 m s⁻¹
• The block does not have negative acceleration.

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🟦 QUESTION 12

The gravitational attraction on the surface of the Moon is about one-sixth of that on the surface of the Earth. An astronaut can throw a ball up to a height of 8 m from the Earth’s surface. How far up will the ball thrown with the same upward velocity travel from the surface of the Moon?

🟩 ANSWER

Maximum height is:$$h=\frac{u^2}{2g}$$h=2gu2​

For the same initial velocity, height is inversely proportional to gravitational acceleration.$$g_{\text{Moon}}=\frac{g_{\text{Earth}}}{6}$$gMoon​=6gEarth​​

Therefore:$$h_{\text{Moon}}=6h_{\text{Earth}}$$hMoon​=6hEarth​$$h_{\text{Moon}}=6\times8$$hMoon​=6×8$$\boxed{h_{\text{Moon}}=48\text{ m}}$$hMoon​=48 m​

The ball will rise to a height of 48 m on the Moon.

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🟦 QUESTION 13

A 1000 kg car is moving along a road at a constant speed. Suddenly, the driver notices an obstruction ahead and applies the brakes to come to a complete stop. The graphical representation of the motion 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

From Fig. 7.38, the speed of the car at A and B is:$$v=35\text{ m s}^{-1}$$v=35 m s−1

(i) Motion between A and B

The graph is horizontal between A and B. Therefore, the car moves with a constant speed of 35 m s⁻¹ and has zero acceleration.

(ii) Kinetic energy at A

$$K=\frac{1}{2}mv^2$$K=21​mv2$$K=\frac{1}{2}\times1000\times35^2$$K=21​×1000×352$$K=500\times1225$$K=500×1225$$\boxed{K=612,500\text{ J}}$$K=612,500 J​

(iii) Work done by brakes from B to C

At C, the car comes to rest.$$K_f=0$$Kf​=0

Using the work-energy theorem:$$W=K_f-K_i$$W=Kf​−Ki​$$W=0-612,500$$W=0−612,500$$\boxed{W=-612,500\text{ J}}$$W=−612,500 J​

The work is negative because the braking force acts opposite to the car’s motion.

(iv) Energy transformation

The kinetic energy of the car is mainly transformed into:

• Heat energy in the brakes
• Heat produced between tyres and the road
• Sound energy
• Internal energy due to deformation

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🟦 QUESTION 14

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−10\text{ m s}^{-1}0 m s−1 and its potential energy is 30 J. Calculate the velocity of the ball at P, Q and R.

🟩 ANSWER

At O:$$K_O=0$$KO​=0$$U_O=30\text{ J}$$UO​=30 J

Therefore, total mechanical energy is:$$E=K_O+U_O=30\text{ J}$$E=KO​+UO​=30 J

Since the track is frictionless, total mechanical energy remains constant.

Velocity at P

From Fig. 7.39:$$U_P=20\text{ J}$$UP​=20 J

Therefore:$$K_P=E-U_P$$KP​=E−UP​$$K_P=30-20=10\text{ J}$$KP​=30−20=10 J

Using:$$K=\frac{1}{2}mv^2$$K=21​mv2$$10=\frac{1}{2}\times0.5\times v_P^2$$10=21​×0.5×vP2​$$10=0.25v_P^2$$10=0.25vP2​$$v_P^2=40$$vP2​=40$$\boxed{v_P=\sqrt{40}\approx6.32\text{ m s}^{-1}}$$vP​=40​≈6.32 m s−1​

Velocity at Q

From the graph:$$U_Q=30\text{ J}$$UQ​=30 J$$K_Q=30-30=0$$KQ​=30−30=0$$\boxed{v_Q=0\text{ m s}^{-1}}$$vQ​=0 m s−1​

Velocity at R

From the graph:$$U_R=40\text{ J}$$UR​=40 J

However, the total energy of the ball is only 30 J.

The ball cannot reach a point where its potential energy alone is greater than its total mechanical energy.$$\boxed{\text{The ball cannot reach point R.}}$$The ball cannot reach point R.​

It stops at Q and reverses its direction.

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🟦 QUESTION 15

A coconut of mass 1.5 kg falls from the top of a coconut tree onto 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. Assume $g=10\text{ m s}^{-2}$g=10 m s−2.

🟩 ANSWER

(i) Velocity before hitting the sand

The coconut falls from rest.$$u=0,\quad g=10\text{ m s}^{-2},\quad h=10\text{ m}$$u=0,g=10 m s−2,h=10 m

Using:$$v^2=u^2+2gh$$v2=u2+2gh$$v^2=0+2\times10\times10$$v2=0+2×10×10$$v^2=200$$v2=200$$v=\sqrt{200}$$v=200​$$\boxed{v=10\sqrt{2}\approx14.14\text{ m s}^{-1}}$$v=102​≈14.14 m s−1​

(ii) Depth of depression

Energy of the coconut just before impact:$$E=mgh$$E=mgh$$E=1.5\times10\times10$$E=1.5×10×10$$E=150\text{ J}$$E=150 J

Work done against the sand:$$W=F\times d$$W=F×d

All the energy is used to create the depression:$$F\times d=150$$F×d=150$$3000d=150$$3000d=150$$d=\frac{150}{3000}$$d=3000150​$$d=0.05\text{ m}$$d=0.05 m$$\boxed{d=5\text{ cm}}$$d=5 cm​

Final Answer:

• Velocity before impact ≈ 14.14 m s⁻¹
• Depth of depression = 0.05 m or 5 cm