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**Work and Energy (Q & A)****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.****Sol.** Here, force acting on the book, F = 40N

distance through which book is displaced, s = 25 cm = 0·25 m

Work done by the force, i.e., W = F × s = (40 N) (0·25 m) = 10J

**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/s ^{2}).**

vertical displacement of the ball, s = 4m

Since the force and the displacement of the ball are in opposite directions, work done by the force of gravity, i.e., W= _F × s = _ (10N) (4m) = _ 40J

Obviously, work done against the force of gravity = 40J

**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.****Sol.** Here, force exerted by the labourer in carrying n bricks (each of mass m kg),

F = (mn) g = (mng) newton

displacement of the bricks, s = h metre

Work done by the labourer, W = F × s = (mng newton) × (h metre) = mngh 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 × 10 ^{5} N.**

distance through which the train moves, s = 1 km = 1000 m

Work done by the frictional force, i.e., W = _ Fs = _ (5 × 10^{5} N) (1000 m) = - 5 × 10^{8} J

(F and s are in opposite directions)

Obviously, work done by the train is 5 × 10^{8} 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/s ^{2}).**

= 80 × 10 N = 800 N (1 kg wt = 10 N)

vertical displacement, s = 100 m

Work done by the man, i.e., W = F × s = (800N) (100m) = 80000 J

**Q6. How fast should a man of mass 60 kg run so that his kinetic energy is 750 J ?****Sol.** Here, mass of the man, m = 60 kg

kinetic energy of the man, E_{k} = 750J

If v is the velocity of the man, then

E_{k} = mv^{2}

or v = = 5 m/s

**Q7. Find the mass of the body which has 5J of kinetic energy while moving at a speed of 2 m/s.****Sol.** Here, kinetic energy of the body, E_{k} = 5J

speed of the body, v = 2 m/s

If m is the mass of the body, then

E_{k} = mv^{2} or = 2.5 kg

**Q8. 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.****Sol.** The ball, which is initially at rest, gains kinetic energy due to work done on it by the player.

Thus, the work done by the player on the ball, W = kinetic energy (E_{k}) of the ball as it leaves his foot, i.e.,

W = E_{k} = mv^{2}

Here, m = 250 g = 0·25 kg, v = 10 m/s

W = (0·25kg) (10 m/s)^{2} = 12·5 J

**Q9. 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 ?****Sol. **Here, mass of the body, m = 5 kg

initial velocity of the body, u = 0

force acting on the body, F = 20 N

time for which the force acts, t = 10 s

If a is the acceleration produced in the body,

a = = 4 m/s^{2}

Let v be the velocity of the body after 10 s.

Clearly, v = u at = 0 (4 m/s^{2}) (10 s) = 40 m/s

Kinetic energy acquired by the body,

E_{k} = mv^{2} = (5kg) (40m/s)^{2} = 4000J

**Q10. 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.****Sol.** Here, mass of the bullet, m = 20 g = 0·02 kg

initial velocity of the bullet, u = 500 m/s

final velocity of the bullet, v = 400 m/s

If W is the work done by the bullet in passing through the tree, then according to work-energy theorem

W = mu^{2} _ mv^{2} = m(u^{2} _ v^{2})

or W = (0·02 kg) [(500 m/s)^{2} _ (400m/s)^{2}] = 900J

**Q11. 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.****Sol.** Here, mass of the body, m = 4 kg

increase in height of the body, h = (10m _ 5m) = 5m

Increase in potential energy, E_{p} = mgh = (4 kg) (10 m/s^{2}) (5m) = 200J

Aliter. Initial potential energy of the body, E_{pi} = mgh = (4 kg) (10 m/s^{2}) (5m) = 200J

Final potential energy of the body, E_{pf} = mgh_{f} = (4 kg) (10 m/s^{2}) (10 cm) = 400J

Increase in potential energy, E_{p} = E_{pf }_ E_{pi} = 400J _ 200J = 200J

**Q12. An object of mass 1 kg is raised through a height h. Its potential energy increases by 1 J. Find the height h.****Sol.** Here, mass of the object, m = 1 kg increase in potential energy, E_{p} = 1J

As E_{p} = mgh, h = = 0·1 m

**Q13. 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.****Sol.** (a) Here, mass of the ball, m = 5 kg,

speed of the ball, v = 10 m/s

Kinetic energy of the ball, E_{k} =mv^{2} = (5kg)(10 m/s)^{2} =250J

When the ball reaches the highest point, Its kinetic energy becomes zero as the entire kinetic energy is converted into its potential energy (E_{p}) i.e., E_{p} = 250J

(b) If h is the maximum height attained by the ball,

E_{p} = mgh

From eqn. (i) and (ii), mgh = 250J

or = 5m

**Q14. 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.****Sol.** Here, mass of the ball, m = 5 kg

Height of the ball, h = 10m

(a) Initial potential energy of the ball,

E_{p} = mgh = (5 kg) (10 m/s^{2}) (10m) = 500J

(b) When the ball reaches the ground, its potential energy becomes zero as it is entirely converted into its kinetic energy (E_{k}), i.e.,

E_{k} = 500J

(c) If v is the velocity attained by the ball before reaching the ground,

or E_{k} = mv^{2}

V= = 14·14 m/s

**Q15. 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.****Sol.** Here, kinetic energy of the body, E_{k} = 10J

maximum height attained by the body, h = 5m

When the body attains maximum height, its entire kinetic energy is converted into its potential energy

(E_{p})' i.e., E_{p} = E_{k}

mgh = 10J

= 0.2 kg

**Q16. A rocket of mass 3 × 10 ^{6} 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.**

velocity acquired by the rocket, v = 1 km/s = 1000 m/s

height attained by the rocket, h = 25 km = 25000 m

(a) Potential energy of the rocket, E_{p} = mgh = (3 × 10^{6} kg) (10 m/s^{2}) (25000 m) =**7·5 × 10 ^{11} J**

(b) Kinetic energy of the rocket,

E_{k} = mv^{2} = (3 × 10^{6} kg) (1000m/s)^{2} = **1·5 × 10 ^{12} J**

**Q17. 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.****Sol.** Here, mass of the boy, m = 40 kg

total height gained, h = 50 × 10 cm = 500 cm = 5m

time taken to climb, t = 5s

Work done by the boy, W = mgh = (40 kg) (10 m/s^{2}) (5m) = 2000J

Power developed, P = = 400W

**Q18. 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 ****Sol.** Here, mass of the coal to be lifted, m = 90 metric tonnes = 90 × 1000 kg=9 × 10^{4} kg

height through which the coal is to be lifted, h = 200m

time during which the coal is to be lifted, t = 1h = 60 × 60 = 3600 s

work done to lift the coal, i.e., W = mgh = (9 × 10^{4} kg) (10 m/s^{2}) (200m) = 18 × 10^{7}J

Power of the engine required i.e., P = = 50000 W = 50 kW

**Q19. How much time does it take to perform 500J of work at a rate of 10W ?****Sol.** Here, work to be performed, W = 500J

rate at which work is to be performed, i.e., power, P = 10W

As P = ,

t = = 50 s

**Q20. Calculate the units of energy consumed by 100 W electric bulb in 5 hours.****Sol.** Here, power of the electric bulb, P = 100 W = 0·1 kW

time for which bulb is used, t = 5h

As P = , W = Pt

Energy consumed by the bulb, W = Pt = 0·1 kW (5 h) = 0·5 kWh = 0·5 units

**Q21. 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.****Sol. **Total distance covered by the lift, s = 10 × 6 m = 60 m

time in which this distance is covered, t = 10 s

force exerted by the lift, F = 4000 kg wt = 4000 × 10 N

= 4 × 10^{4} N (1 kg wt = 10 N)

velocity of the lift, V = = 6 m/ s

Power of the lift, P = F v = (4 × 10^{4} N) (6 m/s) = 24 × 10^{4} W = 240 kW

**Q22. What kind of energy transformation takes place in the following cases ?****(a) When water flowing down a dam runs a turbine to generte electricity.****(b) A running steam engine.****(c) Power generation in a thermal power station.****Ans.** ± The scheme of energy transformation when water stored in a dam is used to produce electricity is,

Potential energy of water ® Kinetic energy of flowing water ® Kinetic energy turbine ® Electricial energy

Thus, in a hydropower station, the potential energy of water sotred in a dam is converted into kinetic energy of the turbine which finally gets converted into electrical energy.

± The scheme of energy transformation in a running steam engine is,

Chemical energy of coal ® Heat energy of steam ® Kinetic energy of moving parts of the engine ® Kinetic energy of the engine and boggies.

So, in a running steam engine the scheme of energy tranformation is,

Chemical energy ® Heat energy ® Kinetic energy

± The scheme of energy transformation in a thermal power stations is,

Chemical energy of fuel (coal / diesel) ® Heat energy of steam ® Kinetic energy of turbine ® Electrical energy

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