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Question

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


A

0.02

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B

0.03

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C

0.06

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D

0.01

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Solution

The correct option is C

0.06


Step 1. Given data:

Mass of marble block, m=2kg

Velocity, v=6ms-1

Stopping time,t=10s

Frictional force, f

Step 2. Calculating back acceleration/Retardation:

Retardation, a=vt

a=610

Therefore, a=0.6m/s2

Step 3. Balancing forces on the marble block

F=ma

f=ma f=μmg

μmg=maμ=agμ=0.610μ=0.06

Hence, option (C) is correct.


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