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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
(d) v=u-at
`0=u-mu gt `
`mu=(u)/(gt)=(6)/(10xx10)=0.06`
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