Laws Of Motion Question 147
Question: A block rests on a rough inclined plane making an angle of $ 30{}^\circ $ with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is $ (takeg=10m/s^{2}) $
Options:
A) 1.6
B) 4.0
C) 2.0
D) 2.5
Show Answer
Answer:
Correct Answer: C
Solution:
[c] The angle of repose, $ \alpha ={{\tan }^{-1}}(\mu )=ta{{n}^{-1}}(0.8)=37{}^\circ $
Here, the angle of inclined plane is $ \theta =37{}^\circ $ .
It means that block is at rest and therefore the friction should be static in nature.
It will balance the component of weight parallel to inclined plane. Static friction=component of weight in downward direction$ =mg\sin \theta =10N $
$ \therefore m=\frac{10}{9\times \sin 30{}^\circ }=2kg $
