A particle moves move on the rough horiz...

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.

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

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

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

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

Text Solution

Verified by Experts

The correct Answer is:

(a) `(3)/(4)`th of KE is lost. Hence left KE is `(1)/(4)`th or,
`v^(2)=(v_(0)^(2))/(4)`
`:. " " v=(v_(0))/(2)=v_(0)-at_(0)=v_(0)-mu" gt"_(0)`
or `" " mu=(v_(0))/(2"gt"_(0))`

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