SATHEE: Laws Of Motion Question 146

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 $θ$ and the coefficient of friction between body and the surface is $μ$ . Then

Options:

A) $t_{1} = t_{2}$

B) $t_{1} > t_{2}$

C) $t_{2} > t_{1}$

D) $t_{1} = 2 t_{2}$

Show Answer

Answer:

Correct Answer: C

Solution:

[c] The retardation r of the block while moving up is $g (s i n θ + μ c o s θ)$ while the acceleration a of the block while moving down is $g (s i n θ - μ \cos θ)$

$t_{1} \propto \frac{1}{\sqrt{r}}$ and $t_{2} \propto \frac{1}{\sqrt{a}}$ since $r = a, t_{2} > t_{1}$

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Laws Of Motion Question 146

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