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\times 9.8\times \sqrt{3}N $

C) $ 9.8\times \sqrt{3}N $

D)$ 0.8\times 9.8N $

Show Answer

Answer:

Correct Answer: A

Solution:

Limiting friction $ F _{l}=\mu \ mg\cos \theta $

$ F _{l}=0.7\times 2\times 10\times \cos 30{}^\circ =12\ N $ (approximately)

But when the block is lying on the inclined plane then component of weight down the plane $ =mg\sin \theta $

$ =2\times 9.8\times \sin 30{}^\circ =9.8\ N $

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



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