# Solve the following :

Question:

A block of mass $2.0 \mathrm{~kg}$ is pushed down an inclined plane of inclination $37^{\circ}$ with a force of $20 \mathrm{~N}$ acting parallel to the incline. It is found that the block moves on the incline with an acceleration of $10 \mathrm{~m} / \mathrm{s}^{2}$. 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 $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$.

Solution:

(a) $S=u t+\frac{1}{2} a t^{2}$

$S=0+\frac{1}{2} a t^{2}$

$\mathrm{S}=5 \mathrm{~m}$

Work done $W=F s \cos \theta=20 \times 5 \times 1$

$\mathrm{W}=100 \mathrm{~J}$

(b) $h=S \sin \theta=S \sin 37^{\circ}=3 m$

work done, $W=m g h=2 \times 10 \times 3$

$\mathrm{W}=60 \mathrm{~J}$

(c) Frictional force $F_{f}=m g \sin \theta$ and work done $\mathrm{W}=\mathrm{F}_{\mathrm{f}} \mathrm{s} \cos \theta$

$W=m g \sin \theta s \sin \theta$

$W=20 \times 0.60 \times 5$

$\mathrm{W}=60 \mathrm{~J}$