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Questions  

In the arrangement shown, coefficient of friction between the block and the surface is 0.6 then 

a
acceleration of the block is 1 m/s2
b
acceleration of the block is 0.5 m/s2
c
frictional force acting on the block is 12 N
d
frictional force acting on the block is 10 N

detailed solution

Correct option is D

fmax⁡=μN=μmg=0.6×2×10 N=12 Nsince 10 N

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