#### Physics (NTSE/Olympiad)

# 3. Work and Energy

__Miscellaneous Examples__

** Example : 18**

A car weighing 1200 kg and travelling at a speed of 20 m/s stops at a distance of 40 m retarding uniformly. Calculate the force exerted by the brakes. Also calculate the work done by the brakes.

** Solution. **In order to calculate the force applied by the brakes, we first calculate the retardation.

Initial speed, u = 20 m/s; final speed,

v = 0, distance covered, s = 90 m

Using the equation, v^{2} = u^{2} + 2as, we get 0^{2} = (20)^{2} + 2 × a × 40

or 80a = –400

or a = –5 m/s^{2}

Force exerted by the brakes is given by

F = ma

Herem = 1200 kg; a = – 5 m/s^{2}

∴ F = 1200 × (–5) = – 6000 N

The negative sign shows that it is a retarding force. Now, the work done by the brakes is given by

W = Fs

Here F = 6000 N; s = 40 m

∴ W = 6000 × 40 J = 240000 J = 2.4 × 10^{5}J

∴ Work done by the brakes = 2.4 × 10^{5}J

** Example : 19**

A horse applying a force of 800 N in pulling a cart with a constant speed of 20 m/s. Calculate the power at which horse is working.

** Solution.** Power, P is given by force × velocity, i.e.

P = F . v

Here, F = 800 N; v = 20 m/s

∴ P = 800 × 20 = 16000 watt = 16 kW

** Example : 20**

A boy keeps on his palm a mass of 0.5 kg. He lifts the palm vertically by a distance of 0.5 m. Calculate the amount of work done.

Use g = 9.8 m/s^{2}.

** Solution. **Work done, W = F . s

Here, force F of gravity applied to lift the mass, is given by

F = mg

= (0.5 kg) × (9.8 m/s^{2})

= 4.9 N

and s = 0.5 m

Therefore, W = (4.9).(0.5m) = 2.45 J.