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 $W=\underset{0}{\overset{2}{\int }}F.dS=\underset{0}{\overset{2}{\int }}ma.dS$

$\underset{0}{\overset{2}{\int }}3\times 2t\times t^{2}dt=\underset{0}{\overset{2}{\int }}6t^{3}dt=\frac{3}{2}[t^4{]}_0^2=24J$