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Question

A particle moves on a rough horizontal ground with some initial velocity v0. If (34)th of its kinetic energy is lost due to friction between the particle and the ground in time interval t0, then find the coefficient of kinetic friction between the particle and ground. (Assume only kinetic friction is acting on the particle)

A
v02gt0
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B
v04gt0
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C
3v04gt0
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D
v0gt0
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Solution

The correct option is A v02gt0
34thof the kinetic energy is lost. Hence final Kinetic energy is 14thof initial.
m2v2=m2v204
v=v02
using kinematics equation
v=v0at0=v0μkgt0
v=v02=v0μkgt0
here a=μkg
μk=v02gt0

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