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A block of mass 2 kg rests on a rough inclined plane making an angle of 300 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
9.8N
b
10N
c
0.98N
d
98N

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detailed solution

Correct option is A

Fl=μmgcosθFl=0.7×2×10×cos30=12N

(approximately) but when the block is lying on the inclined plane then component of weight down the plane = mgsinθ

=2×9.8×sin30=9.8N, It means the body is stationary so static friction will work on it, Static friction = applied force = 9.8N

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