Question

A body of mass $3$kg is under a force, which causes a displacement in it given by $S=\frac{t^3}{3}$ (in m). Find the work done by the force in first $2$ seconds

• A. $24$J
• B. $3.8$J
• C. $5.2$J
• D. $2.6$J

Answer 1

The correct option is A $24$J

Given,

$m=3kg$

$s=\frac{t^3}{3}$

velocity, $v=\frac{ds}{dt}=\frac{d\left(t^3/3\right)}{dt}=t^2$

acceleration, $a=\frac{dv}{dt}=\frac{d\left(t^2\right)}{dt}=2t$

Work done, $W=\int mavdt$

$W={\int }_0^{2}m.2t.t^2.dt$

$W=2m{\int }_0^2{t}^{3}dt=2m\frac{t^4}{4}{|}_0^2$

$W=2\times 3\times \frac{2\times 2\times 2\times 2}{4}=24J$