The displacement x in meter of a particle of mass `mkg` moving in one dimenstion under the action of a force is related to the time t in second by the equation `x=(t-3)^2`. The work done by the force (in joules) in first six seconds is

Options

A. (a) `18m`
B. (b) Zero
C. (c) `9m//2`
D. (d) `36m`

Answer

B

Explanation

Here `x=(t-3)^2=t^2-6t+9`
`v=(dx)/(dt)=2t-6`
At `t=0`, `v=2xx0-6=-6`
At `t=6s`, `v=2xx6-6=+6`
Initial and final KE are same, hence no work is done.