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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 m//s` is stopped by friction in 10 s. 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
Here applying `nu= u + at`
`0 = 6 + axx 10   :. a =-0.6 ms^(-2)`
Also `mu mg = ma` or   `mu=(a)/(g) =( 0.6)/(10) = 0.06`
Hence choice is (d).
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