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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 10s. Then the coefficient of friction is (g = `10 ms^(-2)`)

A

0.02

B

0.03

C

0.06

D

0.01

Text Solution

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The correct Answer is:
C
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A body of mass 2kg lying on ice which given a velocity of 6ms^(-1) is stopped by friction is 5sec find the co-efficient of friction between block and ice (g = 10ms^(-2))