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A marble block of mass 2 kg lying on ice...

A marble block of mass 2 kg lying on ice when given a velocity of `6m//s` is stopped by friction in 10s. Then the coefficient of friction is

A

`0.01`

B

`0.02`

C

`0.03`

D

`0.06`

Text Solution

Verified by Experts

The correct Answer is:
D
`u=6m//s`, `t=10sec`, `u=0`
`v=u-atimpliesa=(6)/(10)=0.6m//s^(2)`
Since, `a=mu g`
`implies0.6=muxx10impliesmu=0.06`.
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