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Q.

A block of mass 2 kg is lying on an inclined plane, inclined to the horizontal at 30°. If the coefficient of friction between the block and the plane is 0.7 then magnitude of frictional force acting on the block will be

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a

11 .9 newton

b

1.19 newton

c

0.19 newton

d

1109 newton

answer is A.

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Detailed Solution

See fig.Here, F=μR=μ⋅mgcos⁡θ=μ⋅mgcos⁡30∘ =0⋅7×2×9⋅8×(3/2)=11⋅9N
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