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Q.

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 mgsin⁡θ=2×9⋅8×12=9⋅8N.
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