Question

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

• A. 2J
• B. 3.8J
• C. 24J
• D. 96J

Answer 1

The correct option is D

96J

$S=\frac{2t^3}{3}\therefore dS=2t^{2}dt;a=\frac{d^{2}S}{d{t}^2}=\frac{d^2}{d{t}^2}\left[2\frac{t^3}{3}\right]=4tm/s^2$

Now work done by the force $W={\int }_0^{2}F.dS={\int }_0^{2}ma.dS$ or ${\int }_0^{2}3\times 4t\times 2t^{2}dt={\int }_0^{2}24t^{3}dt=6{\left[t^4\right]}_0^2=96J$