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Question

# Two blocks of masses 5 kg and 3 kg are placed in contact over an inclined surface of angle 37∘, as shown. μ1 is friction coefficient between 5 kg block and the surface of the incline and similarly, μ2 is friction coefficient between the 3 kg block and the surface of the incline. After the release of the blocks from the inclined surface, (g=10 m/s2)

A
if μ1=0.5 and μ2=0.3, then 5 kg block exerts 3 N force on the 3 kg block
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B
if μ1=0.5 and μ2=0.3, then 5 kg block exerts 8 N force on the 3 kg block
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C
if μ1=0.3 and μ2=0.5, then 5 kg block exerts 1 N force on the 3 kg block
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D
if μ1=0.3 and μ2=0.5, then 5 kg block exerts no force on the 3 kg block
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Solution

## The correct options are A if μ1=0.5 and μ2=0.3, then 5 kg block exerts 3 N force on the 3 kg block D if μ1=0.3 and μ2=0.5, then 5 kg block exerts no force on the 3 kg blockCase-I: μ1=0.5,μ2=0.3 Along the incline, acceleration of 5 kg block will be less than acceleration of 3 kg block provided they move separately on the incline. The reason is greater friction coefficient of 5 g block, as acceleration along the incline is gsinθ−μgcosθ So, both blocks will move together. In this case FBDs of both are shown. For 5 kg block m1gsinθ+N−μ1m1gcosθ=m1a For 3 kg block m2gsinθ−N−μ2m2gcosθ=m2a Solving the two equations, we get N=3 N In second case, acceleration of 5 kg block is more than acceleration of 3 kg. Hence, no normal force will act between them.

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