The displacement x in meter of a particle of mass m kg moving in one dimension 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
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a
18 m
b
Zero
c
9 m/2
d
36 m
answer is B.
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Detailed Solution
Here x=(t−3)2=t2−6t+9 v=dxdt=2t−6At t=0,v=2×0−6=−6At t=6s,v=2×6−6=+6Initial and final KE are same, hence no work is done