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Question

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
0.7×9.8×3 N
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C
9.8×3 N
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D
0.7×9.8 N
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Solution

The correct option is A 9.8 N

Since, block is in rest so applied force will be less than the limiting friction.

μmgcosθ>mgsinθ

Force of friction is f=mgsinθ

f=2×9.8×12=9.8 N

Hence, option (A) is correct.

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