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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
Retardation of the marble block because of friction
`a = (muN)/(m) = (mumg)/(m) =mug`
Now using `v =u + at` we get
`0 =6 - (mu xx 10) xx 10`
Which gives `mu =0.06`
.
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