Q: A block of mass 5 kg slides down on an inclined plane of inclination 30° from a height of 10 m form ground. The coefficient of friction between the block and inclined plane is 0.5. Find the work done by friction on the block.
(a) 250√3 J
(b)-250√3 J
(c) 500 J
(d)-500 J
Ans: (b)
Sol: Frictional force $ f = \mu m g cos \theta $
$ f = 0.5 \times 5 \times 10 cos30°$
$f = 25 \times \frac{\sqrt{3}}{2}$
W = -f × s
Along the inclined plane , s = 20 m
$W = – \frac{25\sqrt{3}}{2}\times 20 $
$ W = – 250\sqrt{3}$ J