Question

A body of mass $4\,kg$ 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 $4\,s$ is,

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

Answer 1

Answer: (B)

The velocity of body at time $t$ is given by,
$v=\frac{d x}{d t}=\frac{d}{d t}\left(t^{2}\right)=2t$ .
Now, at $t=0s$ , initial velocity, $v_{i}=0$ .
At $t=4s$ , final velocity, $v_{f}=8$ .
Work done $=$ increase in $KE$
$=\frac{1}{2}mv_{f}^{2}-\frac{1}{2}mv_{i}^{2}=\frac{1}{2}m\left(v_{t}^{2} - v_{i}^{2}\right)$
$=\frac{1}{2}\times 4\times \left(8^{2} - 0^{2}\right)=128\,J$ .