NCERT Solutions: Work, Energy, and Simple Machines

Table of Contents
1. Think It Over (Page 116)
2. Pause and Ponder (Page 119)
3. Pause and Ponder (Page 121)
4. Pause and Ponder (Page 123)
5. Pause and Ponder (Page 125)
6. Pause and Ponder (Page 129)
7. Pause and Ponder (Page 132)
8. Pause and Ponder (Page 135)
9. What if (Page 135)
10. Revise, Reflect, Refine
View more

Think It Over (Page 116)

Q1: What will be the magnitude of velocity of the child at the bottom of the blue slide?
Ans: The velocity at the bottom depends on the height of the slide and is given by v = √(2gh), where h is the height of the slide.

Q2: Will two children of different masses reach the bottom of the same slide with the same velocity?
Ans: Yes, both will reach with the same velocity because the velocity depends on height and gravity, not on mass.

Q3: Which of the slides will result in the largest magnitude of velocity for the child at its bottom?
Ans: The slide with the greatest height will result in the largest velocity at the bottom, regardless of its shape.

Pause and Ponder (Page 119)

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?

Ans. No. Work done = Force × Displacement. While the weightlifter holds the barbell steady, there is no displacement of the barbell (s = 0). Therefore, work done on the barbell = F × 0 = 0. Even though her muscles use up internal energy (and she feels tired), no scientific work is done on the barbell.

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?

Ans. Negative. The frictional force acts on the stack of coins in a direction opposite to the direction of displacement (friction opposes motion). Since force and displacement are in opposite directions, the work done by friction is negative.

Pause and Ponder (Page 121)

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?

Ans. When pedalling on a flat road at constant speed, muscular (chemical) energy is converted into: (i) Kinetic energy of the bicycle and rider. (ii) Thermal energy due to friction at the wheels, chain, and road surface. (iii) Sound energy (small amounts). (iv) Thermal energy of the surrounding air due to air resistance. On a flat road at constant speed, most energy goes to overcoming friction and air resistance, appearing primarily as thermal energy.

Pause and Ponder (Page 123)

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

Ans.
Kinetic energy = (1/2)mv²

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

Ans. No. Kinetic energy K = ½mv². If velocity is constant, v does not change regardless of position. Since mass m is also constant, kinetic energy remains unchanged as the object moves to different positions.

Pause and Ponder (Page 125)

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?

Ans.

• Horizontal motion at constant velocity: No change in potential energy. U = mgh depends only on height h. Horizontal motion does not change h, so U remains constant.
• Vertical motion (raised upward): Yes, potential energy increases. As h increases, U = mgh increases proportionally. The work done against gravity gets stored as gravitational potential energy.

Pause and Ponder (Page 129)

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.

Ans. Just before hitting the ground (height = 0):

• Potential energy = mg × 0 = 0.
• Using v² = u² + 2as = 0 + 2gh → v² = 2gh.
• Kinetic energy = ½mv² = ½m × 2gh = mgh.
• Total mechanical energy = KE + PE = mgh + 0 = mgh.  This equals the mechanical energy at the top (where KE = 0 and PE = mgh). Hence proved.

Q8. You may have seen a ball roller coaster exhibit (Fig. 7.22). Describe how kinetic 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?

Ans.

• At A (highest point): Maximum potential energy, minimum (nearly zero) kinetic energy.
• At B (lower than A): Some PE converts to KE; both are intermediate values.
• At C (lower than B): Less PE, more KE than at B.
• At bottom points: Minimum PE, maximum KE.

Subsequent peaks D, E are lower because in real life, energy is lost due to friction between the ball and track, and air resistance - this energy is converted to thermal energy. The total mechanical energy decreases progressively, so the ball cannot return to the original height A. Yes, it is entirely due to energy lost because of friction and air resistance.

Pause and Ponder (Page 132)

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

Ans. Roads on hills are built as gentle winding slopes because they act as inclined planes with a larger length L relative to height h. From MA = L/h, a larger L gives a higher MA, meaning less effort (force) is required to drive up the hill. Although the total work done is the same and the distance travelled is more, the force required at any point is much less, making it easier for vehicles and people to climb. A straight vertical road would require a very large force, making it impractical.

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.

Ans. An inclined ladder functions as an inclined plane with MA = L/h, where L is the length of the ladder and h is the height gained. Since L > h, the force required to climb (effort) is less than the person's weight (load). In contrast, a vertical ladder requires a force equal to the person's full body weight at every step. Thus, the inclined ladder requires less effort, making it easier to climb.

Pause and Ponder (Page 135)

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

Ans. When a spoon is used to open a can lid, it acts as a Class I lever - the rim of the can is the fulcrum, effort is applied at the spoon's handle, and load is the lid. The effort arm (handle) is much longer than the load arm. From MA = effort arm / load arm, MA > 1. This means the force applied on the lid is much greater than the force applied by the hand, making it easy to pry open the lid.

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

Ans. Scissors are a Class I lever with fulcrum at the pivot. When the object is placed close to the fulcrum, the load arm d₂ decreases, so MA = d₁/d₂ increases. A higher MA means the cutting force on the hard object is much greater than the effort applied by the hand, making it easier to cut. If the object were placed far from the fulcrum, the load arm would be large, MA would be small, and cutting would be harder.

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

Ans. All real machines involve friction between moving parts. Friction converts mechanical energy into thermal energy (heat). According to the work-energy theorem and conservation of energy, the total energy remains constant - but friction continuously converts useful mechanical energy into thermal energy, which cannot be automatically recovered as mechanical energy. As a result, the mechanical energy of the machine decreases over time and it eventually slows down and stops. A perpetual motion machine would require zero energy loss, which is impossible in reality. Machines only transform energy; they cannot create it.

Revise, Reflect, Refine 

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.

Ans. FALSE. Work = Force × Displacement. If the object does not move (s = 0), work done = 0, regardless of the force applied.

(ii) Lifting a bucket vertically upward results in positive work done on the bucket.

Ans. TRUE. The applied force is upward and the displacement is also upward (same direction). Hence work done = F × s = positive.

(iii) The SI unit for both work and energy is joule (J).

Ans. TRUE. The SI unit of work is joule (J) = 1 N × 1 m. Energy has the same unit as work - joule (J).

(iv) A motionless stretched rubber band has kinetic energy.

Ans. FALSE. A motionless stretched rubber band has elastic potential energy (stored due to deformation), NOT kinetic energy. Since the band is stationary (v = 0), K = ½mv² = 0.

(v) Energy can change from one form to another.

Ans. TRUE. Energy can and does convert from one form to another: electrical energy → light (bulb), chemical energy → mechanical energy (muscles), etc.

Q2. Fill in the blanks.

(i) Work done = ______ × ______ (in the direction of force).

Ans. Work done = Force × Displacement (in the direction of force). W = F × s.

(ii) 1 joule of work is done when a force of ______ newton displaces an object by 1 metre in the direction of the force.

Ans. 1 (one). 1 J = 1 N × 1 m.

(iii) The expression for kinetic energy of a body of mass m and velocity v is ______.

Ans. K = ½mv².

(iv) The potential energy of an object of mass m at a small height h from the Earth's surface is ______.

Ans. U = mgh (where g ≈ 10 m s⁻² near Earth's surface).

(v) Power is defined as the ______ at which work is done.

Ans. Rate. Power is the rate at which work is done: P = W/t.

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.

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

• (i) FALSE - Gravity (mg) always acts on the ball.
• (ii) FALSE - Acceleration = g = 10 m s⁻² downward, not zero.
• (iii) TRUE - At the highest point, velocity = 0, so K = ½mv² = 0.
• (iv) TRUE - Height is maximum at this point, so U = mgh is maximum. Since total ME is conserved and KE = 0, all energy is potential.

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.

Ans:
​(i) Truck moving uphill: Chemical energy (fuel) → Kinetic energy + Potential energy (+ heat energy due to friction)
(ii) Unwinding of a watch spring: Elastic potential energy → Kinetic energy
(iii) Photosynthesis in green leaves: Light energy → Chemical energy
(iv) Water flowing from a dam: Potential energy → Kinetic energy (→ Electrical energy in power plants)
(v) Burning of a matchstick: Chemical energy → Heat and light energy
(vi) Explosion of a firecracker: Chemical energy → Heat, light, sound and kinetic energy
(vii) Speaking into a microphone: Sound energy → Electrical energy
(viii) A glowing electric bulb: Electrical energy → Light and heat energy
(ix) A solar panel: Light energy → Electrical energy

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?

Ans.(i) PE gain in elevator: U = mgh = 50 × 10 × 72.5 = 36250 J. 
(ii) PE gain climbing stairs: U = mgh = 50 × 10 × 72.5 = 36250 J. 
(iii) Conclusion: The gain in potential energy is the same in both cases (36250 J), even though the paths are different. This shows that gravitational potential energy depends only on the height gained (h), not on the path taken. Potential energy is a path-independent quantity.

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.

Ans. Let height of each floor = d. Time to 10th floor = t; time to 20th floor = 2t.

• Energy to 10th floor: E₁ = mg × 10d
• Energy to 20th floor: E₂ = mg × 20d = 2 × E₁ (twice the energy required)
• Power to 10th floor: P₁ = E₁/t
• Power to 20th floor: P₂ = 2E₁/(2t) = E₁/t = P₁ (same power required)

Lifting to the 20th floor requires twice the energy but the same power as lifting to the 10th floor.

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.

Ans.

• Factors determining energy: The energy required = mgh depends on (a) mass of the flag m, (b) acceleration due to gravity g, (c) height of the flagpole h. A fixed pulley only changes the direction of force, not the work required.
• Does speed change work done? No. Work = mgh depends only on mass and height, not on speed. Whether raised slowly or quickly, the same work is done.
• If speed is doubled: Time taken is halved (t/2). New power = mgh / (t/2) = 2 × (mgh/t) = 2P. The power requirement doubles when the speed is doubled.

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.

Ans.

• Day 1: Total mass = 60 + 100 = 160 kg. KE₁ = ½ × 160 × v² = 80v²
• Day 2: Total mass = 60 + 100 + 40 = 200 kg. KE₂ = ½ × 200 × v² = 100v²

Since fuel used ∝ KE: Ratio of fuel used = KE₁ : KE₂ = 80v² : 100v² = 4 : 5

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.

Ans. Using the principle of lever: Effort × effort arm = Load × load arm.

Let weight of child = W, weight of adult = 2W. For balance: W × d_child = 2W × d_adult → d_child = 2 × d_adult.

The child must sit at twice the distance from the fulcrum compared to the adult.

Figure description: Draw a horizontal bar (seesaw) with a pivot (fulcrum △) at the centre. Child (weight W) sits at 2 units from the fulcrum on the left; Adult (weight 2W) sits at 1 unit from the fulcrum on the right. Balance is achieved because W × 2 = 2W × 1.

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)?

Ans.(i) Upward motion: Gravity acts downward, displacement is upward - opposite directions → work done by gravity = negative. Downward motion: Gravity and displacement both act downward - same direction → work done by gravity = positive.

(ii) Initial KE = ½ × 2 × (20)² = 400 J. Without air resistance: h = v²/(2g) = 400/20 = 20 m (ball would reach 20 m). Actual height = 19.4 m. PE at 19.4 m = mgh = 2 × 10 × 19.4 = 388 J. At highest point, final KE = 0. By work-energy theorem: Work by gravity + Work by air resistance = ΔKE = 0 - 400 = -400 J. Work by gravity = -mgh = -388 J. Work by air resistance = -400 - (-388) = -12 J.

Q11. A 10.0 kg block is moving on 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 till 4 m. If the block had a kinetic energy of 180 J 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?

Ans.(i) Speed at 0 m: KE = ½mv². 180 = ½ × 10 × v². v² = 36. v = 6 m s⁻¹.

(ii) Work done by force = Area under force-displacement graph (Fig. 7.37):

• 0 to 2 m (triangle, force 0 to 50 N): Area = ½ × 2 × 50 = 50 J
• 2 to 3 m (rectangle, force = 50 N): Area = 1 × 50 = 50 J
• 3 to 4 m (triangle, force 50 to 0 N): Area = ½ × 1 × 50 = 25 J
• Total work done = 50 + 50 + 25 = 125 J

Final KE = 180 + 125 = 305 J. ½ × 10 × v² = 305 → v² = 61 → v = √61 ≈ 7.81 m s⁻¹.

Negative acceleration: The applied force is always positive (in the direction of motion) and friction is negligible, so no negative acceleration occurs in any portion.

Q12. The gravitational attraction on the surface of the Moon is about 1/6th of that on 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?

Ans. Using conservation of energy: ½mv² = mgh → h = v²/(2g).

• On Earth: h_E = v²/(2g_E) = 8 m
• On Moon: g_M = g_E/6 → h_M = v²/(2 × g_E/6) = 6 × v²/(2g_E) = 6 × 8 = 48 m

The ball will travel 48 m up from the surface of the Moon.

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?

Ans.(i) Between A and B: The car moves at constant speed of 35 m s⁻¹ (uniform motion - this is the driver's reaction time before brakes are applied; no braking force yet).

(ii) KE at A = ½mv² = ½ × 1000 × (35)² = ½ × 1000 × 1225 = 612500 J = 6.125 × 10⁵ J.

(iii) Work done by brakes (B to C): Final KE = 0 (car stops). By work-energy theorem: Work = ΔKE = 0 - 612500 = -612500 J. (Negative because braking force is opposite to displacement.)

(iv) Kinetic energy transforms into: Primarily thermal energy (heat) due to friction between brake pads and wheels, and between tyres and road. A small amount also converts into sound energy.

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.

Ans. Total mechanical energy (conserved on frictionless track) = KE at O + PE at O = 0 + 30 = 30 J.

From Fig. 7.39: PE at P = 20 J; PE at Q = 30 J; PE at R = 40 J.

• At P: KE = 30 - 20 = 10 J. ½ × 0.5 × v² = 10 → v² = 40 → v = √40 ≈ 6.32 m s⁻¹.
• At Q: KE = 30 - 30 = 0 J. v = 0 m s⁻¹. (Ball momentarily stops - a turning point.)
• At R: PE = 40 J > Total ME = 30 J. Since PE > Total ME, KE would need to be negative - which is impossible. Therefore, the ball cannot reach R. It turns back before reaching R.

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.

Ans.(i) Using conservation of energy (PE at top = KE just before impact): mgh = ½mv² → v² = 2gh = 2 × 10 × 10 = 200 m² s⁻²v = √200 = 10√2 ≈ 14.14 m s⁻¹

(ii) KE just before impact = ½mv² = ½ × 1.5 × 200 = 150 J. All this energy is used against the resistive force of sand to create depression of depth d: Work done against sand = Force × depth = 3000 × d = 150 Jd = 150/3000 = 0.05 m = 5 cm