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Question

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=(t3)2. Net work done on the particle (in J) in first six seconds is

A
(12 m) J
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B
Zero
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C
(92)m J
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D
(36 m) J
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Solution

The correct option is B Zero
The displacement is given by x=(t3)2=t26t+9
v=dxdt=2t6,
we know that velocity is the time derivative of displacement.

at t=0, v1=2×06=6
at t=6, v2=2×66=6

K1=12mv21=12m×36=10 m

K2=12mv22=12m×36=10 m

ΔK=K2K1=0

From Work energy theorem, W=ΔK
Hence no work is done.

Hence option B is the correct answer

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