Practice Questions with Solutions: Work & Energy
Q.1. A boy pushes a book by applying a force of 40N. Find the work done by this force as the book is displaced through 25 cm along the path.
Ans: Force acting on the book, F = 40 N
Distance through which the book is displaced, s = 25 cm = 0·25 m
Work done, W = F × s (Here the force acts along the direction of displacement, so cosθ = 1)
W = (40 N) × (0·25 m) = 10 J
Q.2. A ball of mass 1 kg thrown upwards, reaches a maximum height of 4 m. Calculate the work done by the force of gravity during the vertical displacement. (g = 10 m/s2).
Ans: Mass of the ball, m = 1 kg
Acceleration due to gravity, g = 10 m/s2
Force of gravity on the ball, F = mg = (1 kg)(10 m/s2) = 10 N (acting downwards)
Vertical displacement of ball upwards, s = 4 m
Since the force of gravity and the displacement are in opposite directions, the work done by force of gravity is negative:
W = F × s × cos180° = - F s = - (10 N)(4 m) = -40 J
Thus, gravity does -40 J of work while the ball rises; equivalently, 40 J of work is done against gravity.
Q.3. Find the amount of work done by a labourer who carries n bricks of m kilogram each to the roof of a house h metre high by climbing a ladder.
Ans: Total mass being lifted = n × m = mn kg
Force exerted to lift the bricks (weight), F = mn g newton
Vertical displacement, s = h metre
Work done, W = F × s = (mn g) × h = m n g h joule
Q.4. An engine pulls a train 1 km over a level track. Calculate the work done by the train given that the frictional resistance is 5 × 105 N.
Ans: Frictional resistance acting opposite to motion, F = 5 × 105 N
Distance moved, s = 1 km = 1000 m
Work done by the frictional force = - F s = - (5 × 105 N)(1000 m) = -5 × 108 J (negative because friction opposes motion)
To overcome friction, the engine must do positive work of magnitude 5 × 108 J. Thus, work done by the engine (or by the train) = +5 × 108 J.
Q.5. A man weighing 70 kg carries a weight of 10 kg on the top of a tower 100 m high. Calculate the work done by the man. (g = 10 m/s2).
Ans: Interpretation: The man climbs to the top carrying the 10 kg load. If we consider the work done by the man in lifting both his own body and the extra weight, total mass lifted = 70 + 10 = 80 kg.
Total weight (force) = 80 kg × 10 m/s2 = 800 N
Vertical displacement, s = 100 m
Work done, W = F × s = (800 N)(100 m) = 80 000 J
If only the 10 kg weight were considered, the work would be (10 × 10 × 100) = 10 000 J. The answer above follows the stated assumption of lifting both masses.
Q.6. How fast should a man of mass 60 kg run so that his kinetic energy is 750 J ?
Ans: Mass, m = 60 kg
Kinetic energy required, Ek = 750 J
Use Ek = 1/2 m v2
So v = sqrt(2 Ek / m)
v = sqrt(2 × 750 J / 60 kg) = sqrt(1500 / 60) = sqrt(25) = 5 m/s
Thus, the man should run at 5 m/s.
or
v= 
= 5 m/s
Q.7. Find the mass of the body which has 5J of kinetic energy while moving at a speed of 2 m/s.
Ans: Kinetic energy, Ek = 5 J
Speed, v = 2 m/s
Use Ek = 1/2 m v2
So m = (2 Ek) / v2 = (2 × 5 J) / (2 m/s)2 = 10 / 4 = 2·5 kg
Q.8. A player kicks a ball of mass 250 g at the centre of a field. The ball leaves his foot with a speed of 10 m/s, Find the work done by the player on the ball.
Ans: Mass of the ball, m = 250 g = 0·25 kg
Speed after kick, v = 10 m/s
Work done by the player = increase in kinetic energy = Ek = 1/2 m v2
W = 1/2 × 0·25 kg × (10 m/s)2 = 0·125 × 100 = 12·5 J
Thus, the player does 12·5 J of work on the ball.
Q.9. A body of mass 5 kg, initially at rest, is subjected to a force of 20N. What is the kinetic energy acquired by the body at the end of 10s ?
Ans: Mass, m = 5 kg; initial velocity, u = 0; Force, F = 20 N; time, t = 10 s
Acceleration, a = F / m = 20 N / 5 kg = 4 m/s2
Velocity after 10 s, v = u + a t = 0 + (4 m/s2)(10 s) = 40 m/s
Kinetic energy, Ek = 1/2 m v2 = 1/2 × 5 kg × (40 m/s)2 = 0·5 × 5 × 1600 = 4000 J
Q.10. A bullet of mass 20 g moving with a velocity of 500 m/s, strikes a tree and goes out from the other side with a velocity of 400 m/s. Calculate the work done by the bullet in joule in passing through the tree.
Ans: Mass of the bullet, m = 20 g = 0·02 kg
Initial speed, u = 500 m/s; Final speed, v = 400 m/s
Change in kinetic energy of the bullet, ΔE = Ek,f - Ek,i = 1/2 m v2 - 1/2 m u2 = 1/2 m (v2 - u2)
ΔE = 1/2 × 0·02 kg × (4002 - 5002) = 0·01 × (160000 - 250000) = 0·01 × (-90000) = -900 J
Hence, the bullet does 900 J of work on the tree.
Q.11. A body of mass 4 kg is taken from a height of 5 m to a height 10 m. Find the increase in potential energy.
Ans: Mass, m = 4 kg
Initial height = 5 m; Final height = 10 m; Increase in height, Δh = 10 m - 5 m = 5 m
Increase in potential energy, ΔEp = m g Δh = (4 kg)(10 m/s2)(5 m) = 200 J
Alternatively: Initial Ep,i = m g hi = (4)(10)(5) = 200 J; Final Ep,f = (4)(10)(10) = 400 J; Increase = 400 J - 200 J = 200 J.
Q.12. An object of mass 1 kg is raised through a height h. Its potential energy increases by 1 J. Find the height h.
Ans: Mass, m = 1 kg; Increase in potential energy, ΔEp = 1 J
Use ΔEp = m g h ⇒ h = ΔEp / (m g) = 1 J / (1 kg × 10 m/s2) = 0·1 m
Q.13. A 5 kg ball is thrown upwards with a speed of 10 m/s.
(a) Find the potential energy when it reaches the highest point.
(b) Calculate the maximum height attained by it.
Ans: (a) Mass, m = 5 kg; Initial speed, v = 10 m/s
Initial kinetic energy, Ek = 1/2 m v2 = 1/2 × 5 × (10)2 = 0·5 × 5 × 100 = 250 J
At the highest point kinetic energy becomes zero and is converted into potential energy. Therefore potential energy at highest point = 250 J.
(b) Let h be the maximum height. Then m g h = 250 J
= 5mQ.14. A 5 kg ball is dropped from a height of 10m.
(a) Find the initial potential energy of the ball.
(b) Find the kinetic energy just before it reaches the ground and
(c) Calculate the velocity before it reaches the ground.
Ans: Mass, m = 5 kg; Height, h = 10 m; g = 10 m/s2
(a) Initial potential energy, Ep = m g h = (5 kg)(10 m/s2)(10 m) = 500 J
(b) Just before reaching the ground all potential energy is converted into kinetic energy (neglecting air resistance). So Ek = 500 J.
(c) If v is the speed just before impact, Ek = 1/2 m v2
= 14·14 m/sQ.15. A body is thrown up with a kinetic energy of 10 J. If it attains a maximum height of 5 m, find the mass of the body.
Ans: Given initial kinetic energy = 10 J; At maximum height it is entirely converted into potential energy, so m g h = 10 J
Given h = 5 m and g = 10 m/s2
m =
= 0.2 kgQ.16. A rocket of mass 3 × 106 kg takes off from a launching pad and acquires a vertical velocity of 1 km/s and an altitude of 25 km. Calculate its (a) potential energy (b) kinetic energy.
Ans: Mass, m = 3 × 106 kg
Velocity, v = 1 km/s = 1000 m/s
Height, h = 25 km = 25 000 m; g = 10 m/s2
(a) Potential energy, Ep = m g h = (3 × 106)(10)(25 000) = 3 × 106 × 250 000
= 7·5 × 1011 J
(b) Kinetic energy, Ek = 1/2 m v2 = 1/2 × (3 × 106) × (1000)2 = 0·5 × 3 × 106 × 106
= 1·5 × 1012 J
Q.17. A boy of mass 40 kg runs up a flight of 50 steps, each of 10 cm high, in 5 s. Find the power developed by the boy.
Ans: Mass, m = 40 kg
Total height gained = 50 × 10 cm = 500 cm = 5 m
Time taken, t = 5 s
Work done (gain in potential energy) = m g h = (40)(10)(5) = 2000 J
Power P = W / t = 2000 J / 5 s = 400 W
Thus, the boy develops 400 W of power while climbing.
Q.18. What should be the power of an engine required to lift 90 metric tonnes of coal per hour from a mine whose depth is 200 m
Ans: Mass of coal per hour, m = 90 metric tonnes = 90 × 1000 kg = 9 × 104 kg
Height to lift, h = 200 m; Time, t = 1 hour = 3600 s; g = 10 m/s2
Work done per hour, W = m g h = (9 × 104)(10)(200) = 1·8 × 108 J
Power required, P = W / t = 1·8 × 108 J / 3600 s = 5 × 104 W = 50 kW
Q.19. How much time does it take to perform 500J of work at a rate of 10W ?
Ans: Work, W = 500 J; Power, P = 10 W
Time, t = W / P = 500 J / 10 W = 50 s
Q.20. Calculate the units of energy consumed by 100 W electric bulb in 5 hours.
Ans: Power of bulb, P = 100 W = 0·1 kW; Time, t = 5 h
Energy consumed, E = P t = 0·1 kW × 5 h = 0·5 kWh = 0·5 units
Q.21. A lift is designed to carry a load of 4000 kg through 10 floors of a building, averaging 6 m per floor, in 10 s. Calculate the power of the lift.
Ans: Total distance (height) s = 10 × 6 m = 60 m; Time, t = 10 s
Load mass = 4000 kg; Force to lift = weight = 4000 × 10 N = 4 × 104 N
Average speed, v = s / t = 60 m / 10 s = 6 m/s
Power required, P = F v = (4 × 104 N)(6 m/s) = 24 × 104 W = 240 kW
Q.22. What kind of energy transformation takes place in the following cases?
(a) When water flowing down a dam runs a turbine to generate electricity.
(b) A running steam engine.
(c) Power generation in a thermal power station.
Ans:
(a) The energy transformation is: Potential energy of stored water → Kinetic energy of flowing water → Mechanical energy of the turbine → Electrical energy.
(b) Energy transformation in a running steam engine is: Chemical energy of fuel (coal) → Heat energy (steam) → Mechanical energy of moving parts → Mechanical work done by the engine.
(c) In a thermal power station the scheme is: Chemical energy of fuel (coal/diesel) → Heat energy (steam) → Mechanical energy of turbine → Electrical energy produced by the generator.
FAQs on Practice Questions with Solutions: Work & Energy
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