Solve the following :

Question:
A block of mass $2.0 \mathrm{~kg}$ kept at rest on an inclined plane of inclination $37^{\circ}$ is pulled up the plane by applying a constant force of $20 \mathrm{~N}$ parallel to the incline. The force acts for one second. (a) Show that the work done by the applied force does not exceed $40 \mathrm{~J}$. (b) Find the work done by the force of gravity in that one second if the work done by the applied force is $40 \mathrm{~J}$. (c) Find the kinetic energy of the block at the instant the force ceases to act. Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$.

Solution:

(a) For a free body,

Force $F=m g \sin \theta+m a$

$\mathrm{a}=\frac{(\mathrm{F}-\mathrm{mg} \sin \theta)}{m}$

$\mathrm{a}=\frac{\left(20-2 \times 10 \times \sin 37^{2}\right)}{2}$

$a=4 \mathrm{~m} / \mathrm{s}^{2}$

$s=u t+\frac{1}{2} a t^{2}$

$s=0 \times 1+\frac{1}{2} \times 4 \times 1^{2}$

$s=2 m$

$W=F s=20 \times 2$.

$W=40 \mathrm{~J}$

(b) $W=F s$

$\mathrm{s}=\bar{F}=\frac{40}{20}$

$s=2$

$W=-m g h=-m g\left[s \sin 37^{\circ}\right]$

$W=-20 \times 1.2$

$W=-24 \mathrm{~J}$

$v=40 \mathrm{~m} / \mathrm{s}$

$K \cdot E=\frac{1}{2} m v^{2}$

K.E. $=16 \mathrm{~J}$

Solve the following :

Leave a comment

All Study Material