Find the minimum value of coefficient of friction between the 4kg block and the surface for the system to be at rest for the figure shown, (Block A=4kg and block B=3kg)
A
0.4
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B
0.5
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C
0.6
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D
0.75
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Solution
The correct option is D0.75 By FBD of the blocks
For the system to be in rest, a=0
So, T=3g
From FBD of 4kg block f=T, where f=μN is the frictional force on the block
On equating both the equation 4μg=3g μ=34=0.75