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
BETA
