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
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a
9.8 N
b
0.7 × 9.8 × √3 N
c
9.8 × √3 N
d
0.7 × 9.8 N
answer is A.
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Detailed Solution
Angle of friction φ = tan-10.7 >300 . Therefore the block will be in eqilibrium .The actual frictional force will be equal to mgsinθ=2×9⋅8×12=9⋅8N.