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Question

A block of mass 2 kg rests on a rough inclined plane making an angle of 30 with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block is

A
9.8 N
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B
9.83 N
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C
4.93 N
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D
19.6 N
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Solution

The correct option is A 9.8 N
Angle of repose θr=tan1(μs)=tan1(0.7)
or tan θr=0.7
Angle of plane is θ=30,tanθ=tan30=0.577

Since tanθ<tanθr
θ<θr
Block will not slide
f=mg sinθμmg cosθ
or f=2×9.8×sin30=9.8 N

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