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Question

# A particle moves on rough horizontal ground with some initial velocity v0. if 3/4 of its kinetic energy is lost in friction in time t0 the coefficient of friction between the particle and ground.

A
v02gto
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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 C v02gtoInitial velocity=v0As 3/4 th kinetic energy lost so 1/4 kinetic energy remain after time 't0'So velocity after time 't0'= v0/2retardation due to friction=−μgapply first equation of motionv=u+atv0/2=v0−μgt0μ=v0/2gt0

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