Question

A block is placed at the bottom of an inclined plane and projected upwards with some initial speed. It slides up the incline, stops after time $t_1$, and slides back in a further time $t_2$. The angle of inclination of the plane with the horizontal is $\theta$ and the coefficient of friction is $\mu$. Then,

• A. $t_1>t_2$
• B. $t_1<t_2$
• C. the retardation of the block while moving up is $g(\sin \theta +\mu \cos \theta )$.
• D. the acceleration of the block while moving down is $g(\sin \theta -\mu \cos \theta )$.

Answer 1

Answer: (B), (C), (D)

The correct options are
B the retardation of the block while moving up is $g(\sin \theta +\mu \cos \theta )$.
C $t_1<t_2$
D the acceleration of the block while moving down is $g(\sin \theta -\mu \cos \theta )$.

Here, a block is placed at the bottom of an inclined plane and projected upwards with some initial speed.
For the upward motion
$ma=mg\sin \theta +F$
$ma=mg\sin \theta +\mu mg\cos \theta$
Therefore, retardation is
$a=g\sin \theta +\mu g\cos \theta$
$a=g(\sin \theta +\mu \cos \theta )$...........(1)
The block slides up the incline and stops after time $t_1$ and slides back in a further time $t_2$.
For the downward motion
$ma=mg\sin \theta -F$
$ma=mg\sin \theta -\mu mg\cos \theta$
Therefore, acceleration is
$a=g\sin \theta -\mu g\cos \theta$
$a=g(\sin \theta -\mu \cos \theta )$...........(2)
Now, we know
$v=u+at$
$\Rightarrow t=\frac{v-u}{a}$
$\Rightarrow t\propto \frac{1}{a}$
Now from (1) and (2) it is clear that, (1) > (2) hence,
$t_1<t_2$