Question

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

• A. 2 J
• B. 3.8 J
• C. 5.2 J
• D. 24 J

Answer 1

The correct option is D 24 J

$S=\frac{t^3}{3}\therefore dS=t^{2}dt$
$a=\frac{d^{2}S}{d{t}^2}=\frac{d^2}{d{t}^2}[\frac{t^3}{3}]=2tm/s^2$
Now work done by the force $W={\int }_0^{2}F.dS={\int }_0^{2}ma.dS$
${\int }_0^{2}3\times 2t\times t^{2}dt={\int }_0^{2}6t^{3}dt=\frac{3}{2}[t^4{]}_0^2=24J$