Chapter : 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, v2 = u2 + 2as, we get 02 = (20)2 + 2 × a × 40
or 80a = –400
or a = –5 m/s2
Force exerted by the brakes is given by
F = ma
Herem = 1200 kg; a = – 5 m/s2
∴ 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 × 105J
∴ Work done by the brakes = 2.4 × 105J
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/s2.
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/s2)
= 4.9 N
and s = 0.5 m
Therefore, W = (4.9).(0.5m) = 2.45 J.
Trending Articles & Blogs
Download Old Sample Papers For Class X & XII
Download Practical Solutions of Chemistry and Physics for Class 12 with Solutions