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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 v02gto
Initial velocity=v0
As 3/4 th kinetic energy lost so 1/4 kinetic energy remain after time 't0'
So velocity after time 't0'= v0/2
retardation due to friction=μg
apply first equation of motion
v=u+at
v0/2=v0μgt0
μ=v0/2gt0

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