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
9.8√3 N
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C
4.9√3 N
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D
19.6 N
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Solution
The correct option is A9.8 N Angle of repose θr=tan−1(μs)=tan−1(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 sinθ≠μmg cosθ
or f=2×9.8×sin30∘=9.8 N