A particle of mass 0.01kg travels with a velocity given by 4^i+16^kms−1. After sometime, its velocity becomes 8^i+20^jms−1. The work done on particle during this interval of time is
A
0.32J
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B
6.9J
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C
9.6J
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D
0.96J
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Solution
The correct option is D0.96J Magnitude of initial velocity of particle is, vi=√42+162=√272m/s
Magnitude of final velocity of particle is, vf=√82+202=√464m/s
From work energy theorem, Wext=ΔKE Wext=12m(v2f−v2i) Wext=12×0.01×(464−272) ∴Wext=0.96J