Question

Under the action of a force, a $2$ kg body moves such that its position $x$ as a function of time t 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 two seconds is:

• A. 1600 J
• B. 160 J
• C. 16 J
• D. 1.6 J

Answer 1

The correct option is C 16 J

$v=\frac{dx}{dt}=t^2$

$W=\frac{1}{2}m{v}^2=\frac{1}{2}m{t}^4$

$=\frac{1}{2}\times 2\times (2{)}^4=16J$