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Question

# A particle moves on the rough horizontal ground with some initial velocity say v. If 3/4th of its kinetic energy is lost in friction in time t then coefficient of friction between the particle and the ground is:

A
v/2gt
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B
v/4gt
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C
3v/4gt
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D
v/gt
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Solution

## The correct option is A v/2gtInitial velocity=vAs 3/4 th kinetic energy lost so 1/4 kinetic energy remain after time 't'So velocity after time 't'= v/2retardation due to friction=−μgapply first equation of motionv=u+atv/2=v−μgtμ=v/2gt

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