Question

A body of mass $4kg$ is acted upon by a force which causes a displacement in it given by, $x=t^{2}m$ , where $t$ is time in second. The work done by force in $4s$ is,

• A. $64J$
• B. $128J$
• C. $200J$
• D. $240J$

Answer 1

Answer: (B)

The velocity of body at time $t$ is given by,
$v=\frac{dx}{dt}=\frac{d}{dt}\left(t^2\right)=2t$ .
Now, at $t=0s$ , initial velocity, $v_i=0$ .
At $t=4s$ , final velocity, $v_f=8$ .
Work done $=$ increase in $KE$
$=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2=\frac{1}{2}m\left(v_t^2-v_i^2\right)$
$=\frac{1}{2}\times 4\times \left(8^2-0^2\right)=128J$ .