The displacement x (in meter) of a particle of mass m(in kg) varies with time t (in seconds) as given by the equation x=(t−3)2. Net work done on the particle (in J) in first six seconds is
A
(12m)J
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B
Zero
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C
(92)mJ
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D
(36m)J
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Solution
The correct option is B Zero The displacement is given by x=(t−3)2=t2−6t+9 v=dxdt=2t−6,
we know that velocity is the time derivative of displacement.
at t=0,v1=2×0−6=−6
at t=6,v2=2×6−6=6
K1=12mv21=12m×36=10m
K2=12mv22=12m×36=10m
ΔK=K2−K1=0
From Work energy theorem, W=ΔK
Hence no work is done.