Question
A body of mass $$3\,kg$$ is under a constant force, which causes a displacement $$s$$ in metre in it, given by the relation $$s = \frac{1}{3}{t^2},$$ where $$t$$ is in second. Work done by the force in $$2\,s$$ is
A.
$$\frac{5}{{19}}\,J$$
B.
$$\frac{3}{8}\,J$$
C.
$$\frac{8}{3}\,J$$
D.
$$\frac{{19}}{5}\,J$$
Answer :
$$\frac{8}{3}\,J$$
Solution :
Work done by the force $$ = {\text{force}} \times {\text{displacement}}$$
$${\text{or}}\,\,W = F \times s\,......\left( {\text{i}} \right)$$
But from Newton's 2nd law, we have
$${\text{Force}} = {\text{mass}} \times {\text{acceleration}}$$
$${\text{i}}{\text{.e}}{\text{.}}\,\,F = ma\,......\left( {{\text{ii}}} \right)$$
Hence, from Eqs. (i) and (ii), we get
$$W = mas = m\left( {\frac{{{d^2}s}}{{d{t^2}}}} \right)s\left( {\because a = \frac{{{d^2}s}}{{d{t^2}}}} \right)\,\,......\left( {{\text{iii}}} \right)$$
Now, we have, $$s = \frac{1}{3}{t^2}$$
$$\eqalign{
& \therefore \frac{{{d^2}s}}{{d{t^2}}} = \frac{d}{{dt}}\left[ {\frac{d}{{dt}}\left( {\frac{1}{3}{t^2}} \right)} \right] \cr
& = \frac{d}{{dt}} \times \left( {\frac{2}{3}t} \right) = \frac{2}{3}\frac{{dt}}{{dt}} = \frac{2}{3} \cr} $$
Hence, Eq. (iii) becomes
$$W = \frac{2}{3}ms = \frac{2}{3}m \times \frac{1}{3}{t^2} = \frac{2}{9}m{t^2}$$
We have, $$m = 3\,kg,t = 2\,s$$
$$\therefore W = \frac{2}{9} \times 3 \times {\left( 2 \right)^2} = \frac{8}{3}J$$