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Question

A car begins to move at time t=0 and then accelerates along a straight track with a speed given by V(t)=2t2 ms1 for 0t2.
After the end of acceleration, the car continues to move at constant speed. A small block initially at rest on the floor of the car begins to slip at t=1sec and stops slipping at t=3sec. Find the coefficient of static and kinetic friction between the block and the floor.

A
μs=0.4,μk=0.3
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B
μs=0.3,μk=0.4
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C
μs=0.2,μk=0.5
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D
μs=0.5,μk=0.2
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Solution

The correct option is C μs=0.4,μk=0.3
a(t)=dv(t)dt=4t
At t=1, a=4 m/s2
When the block is just starting to slip we have,
μsN=ma
μsmg=4m
μs=0.4
Now it goes on slipping till 3 seconds, therefore in 2 seconds the block's velocity changes from 2to2×(2)2=8m/sec
Hence, acceleration from t=1 to t=3 secs is: aavg=822=3m/s2
Applying equation of motion
μk×m×g=m×aavg=3m
μk=0.3

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