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A particle moves on a rough horizontal g...

A particle moves on a rough horizontal ground with some initial velocity say `nu_(0)`. If (3/4)th of its kinetic energy is lost in friction in time `t_(0)`, then coefficient of friction between the particle and the ground is:

A

`nu_(0)/(2"gt"_(0))`

B

`nu_(0)/(4"gt"_(0))`

C

`(3nu_(0))/(4"gt"_(0))`

D

`nu_(0)/"gt"_(0)`

Text Solution

Verified by Experts

The correct Answer is:
a
`3//4^(th)` energy is lost.i,e., `1/4 ^(th)` kinetic energy is left. Hance, its velocity become `nu_(0)/2` under a retardation ofnmmg in time `t_(0)`.
`nu_(0)/2=nu_(0)-mu"gt"_(0) or mu=nu_(0)/(2"gt"_(0))`
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Knowledge Check

  • A particle moves move on the rough horizontal ground with some initial velocity V_(0) . If (3)/(4) of its kinetic enegry lost due to friction in time t_(0) . The coefficient of friction between the particle and the ground is.

    A
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    A
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