Question

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

• A. 2 J
• B. 3.8 J
• C. 24 J
• D. 96 J

Answer 1

The correct option is D 96 J

$S=\frac{2t^3}{3}\therefore dS=2t^{2}dta=\frac{d^{2}S}{d{t}^2}=\frac{d^2}{d{t}^2}\left[2\frac{t^3}{3}\right]=4tm/s^2$
Now work done by the force $W={\int }_0^{2}FdS={\int }_0^{2}madS$
${\int }_0^{2}3\times 4t\times 2t^{2}dt={\int }_0^{2}24t^{3}dt=6[t^4{]}_0^2=96J$