Question

Under the action of a force, a $2\text{kg}$ 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 meter and $t$ in second. The work done by the force in the first two seconds is

• A. $1600\text{J}$
• B. $160\text{J}$
• C. $16\text{J}$
• D. $1.6\text{J}$

Answer 1

The correct option is C $16\text{J}$

$v=\frac{dx}{dt}=t^2$
At $t=0,v=0$
At $t=2s,v=4\text{m/s}$
From work energy theorem.
W = change in kinetic energy
$=K_f-K_i$
$=\frac{1}{2}m(v_f^2-v_1^2)$
$=\frac{1}{2}\times 2\times (16-0)$
= $16\text{J}$