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⋅mgcosθ=μ⋅mgcos30∘ =0⋅7×2×9⋅8×(3/2)=11⋅9N