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Question

# A marble block of mass 2 kg lying on the ice, when given a velocity of 6 m/s is stopped by friction in 10 s. Then find the value of the coefficient of friction if g=10 m/s2.

A
0.01
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B
0.02
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C
0.03
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D
0.06
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Solution

## The correct option is D 0.06The given situation can be reframed as shown below Let the coefficient of the friction be μ Therefore, friction force acting on block, f=μmg=2μg So, from the FBD we have a=fm=−2μg2=−μg Since, retardation takes place here, we have taken acceleration a as −ve So, from the equation of motion we have v=u+at Since the blocks finally comes to rest, ⇒ 0=u−μgt ∴ μ=ugt=610×10=0.06

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