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The displacement x in meter of a particl...

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

A

(a) `18m`

B

(b) Zero

C

(c) `9m//2`

D

(d) `36m`

Text Solution

Verified by Experts

The correct Answer is:
B
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.
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