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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 v02gt034thof the kinetic energy is lost. Hence final Kinetic energy is 14thof initial. m2v2=m2v204 ∴ v=v02 using kinematics equation v=v0−at0=v0−μkgt0 ∴ v=v02=v0−μkgt0 here a=μkg μk=v02gt0

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