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Question

# A particle moves on a rough horizontal ground with some initial velocity say v0. If 34 of its kinetic energy is lost due to friction in time t0 then coefficient of friction between the particle and the ground is :

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 v02gt0Initial K.EK.Ei=mv202Energy loss in friction is the work done by the frictionWorkdone by frictionW=(34)mv202=3mv208Speed be vmv22=mv208v=v02Let the coefficient be μa=−μg−μg=v02−v0t0μg=v02t0μ=v02gt0

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