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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 `6ms^(-1)` is stopped by friction in 10 s. Then the coefficient of friction is `(g = 10ms^(-2))`

A

0.02

B

0.03

C

0.06

D

0.01

Text Solution

Verified by Experts

The correct Answer is:
C
`v=u+at, a=mu_(k)g`
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