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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 6 `ms^(-1)` is stopped by friction is 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 f
`a = (mu N)/(m)=(mu m g )/(m) =mu g `
Now using V =u + at, we get,
`0 = 6 - (mu xx 10 )xx10 `
Which gives `mu = 0.06`
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