Laws Of Motion Question 58

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[IIT 1980; J & K CET 2004]

Options:

A) 9.8 N

B) 0.7×9.8×3N

C) 9.8×3N

D)0.8×9.8N

Show Answer

Answer:

Correct Answer: A

Solution:

Limiting friction Fl=μ mgcosθ

Fl=0.7×2×10×cos30=12 N (approximately)

But when the block is lying on the inclined plane then component of weight down the plane =mgsinθ

=2×9.8×sin30=9.8 N

It means the body is stationary, so static friction will work on it Static friction = Applied force = 9.8 N



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