Laws Of Motion Question 146
Question: A block is placed at the bottom of an inclined plane and projected upwards with some initial speed. It slides up the plane and stops after time $ t _{1} $ . It begins to slide back down to the bottom in a further time $ t _{2} $ . The angle of inclination of plane is $ \theta $ and the coefficient of friction between body and the surface is $ \mu $ . Then
Options:
A) $ t _{1}=t _{2} $
B) $ t _{1}>t _{2} $
C) $ t _{2}>t _{1} $
D) $ t _{1}=2t _{2} $
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Answer:
Correct Answer: C
Solution:
[c] The retardation r of the block while moving up is $ g(sin\theta +\mu cos\theta ) $ while the acceleration a of the block while moving down is $ g(sin\theta -\mu \cos \theta ) $
$ t _{1}\propto \frac{1}{\sqrt{r}} $ and $ t _{2}\propto \frac{1}{\sqrt{a}} $ since $ r=a,t _{2}>t _{1} $
BETA
