Question

Under the action of a force, a $2kg$ body moves such that its position $x$ as a function of time is given by $x=\frac{t^3}{3}$ where $x$ is in metre and $t$ in second. The work done by the force in the first $2s$ is :

• A. $1,600J$
• B. $160J$
• C. $16J$
• D. $1.6J$

Answer 1

Answer: (C)

$16J$

Position of the body varies as $x=\frac{t^3}{3}$

Thus the acceleration of the body at any instant=$\frac{d^{2}x}{d{t}^2}=2t$

Force any on body at any instant=$ma=2mt$

Thus work done on body in moving the object by a distance $dx=F.dx=2mtdx$

$=2mt\left(t^{2}dt\right)$

Thus total work done in $2s$=${\int }_0^{2}2m{t}^{3}dt=16J$