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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

`0.02`

B

`0.03`

C

`0.06`

D

`0.01`

Text Solution

Verified by Experts

The correct Answer is:
C
Let coefficient of friction be µ, and then retardation will be `mu` g.
From equation of motion, v = u + at
`implies 0 = 6 - mu g xx 10`
`implies mu = (6)/(100) =0.06 `
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