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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/2gt
Initial velocity=v
As 3/4 th kinetic energy lost so 1/4 kinetic energy remain after time 't'
So velocity after time 't'= v/2
retardation due to friction=μg
apply first equation of motion
v=u+at
v/2=vμgt
μ=v/2gt

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