A block of mass 2.0 kg is pushed down an inclined plane of inclination 37° with a force of 20 N acting parallel to the incline. It is found that the block moves on the incline with an acceleration of 10 m/s2. If the block started from rest, find the work done (a) by the applied force in the first second, (b) by the weight of the block in the first second and (c) by the frictional force acting on the block in the first second. Take g = 10 m/s^{2}.

#### Solution

Given:

\[\text{ Mass, m = 2 kg } \]

\[\text{ Inclination, } \theta = 37^\circ\]

\[\text{ Force applied, F = 20 N } \]

\[\text{ Acceleration of the block, a = 10 m/ s}^2\]

(a) t = 1 sec

So,

\[s = \text{ ut } + \frac{1}{2}\text{ at }^2 = 5 \text{m } \]

Work done by the applied force,

\[\text{ W = mgh} \]

\[2 \times 10 \times 3 = 60 J\]

So, frictional force,

\[\text{ W = fs } \cos 0^\circ= \left( \text{ mg } \sin \theta \right) s\]

\[ = 20 \times 0 . 60 \times 5 = - 60 J\]