Question

A belt is moving with constant speed u (clockwise, as shown in figure) on two fixed wheels. The length of the belt between $\ell =10m$, as shown. At time t = 0, a small block starts moving from the upper right end of the belt with leftward speed v (w. r. t. ground). $\mu$ is friction coefficient between belt and block. The moving part of the belt between the wheels is always horizontal.

Answer 1

In all the events, going to the left side is taken as positive and to the right as negative.
A.
$mg\mu =ma$

hence, $a=\mu g$ = 1 $m/s^2$

Writing the equation of motion as displacement being 10 m, to calculate time,
$10=\frac{21}{2}t-\frac{1}{2}\left(1\right)t^2$
This gives t = 1 s, and 20 s.
We know that after 1 s, the block will fall off the left end. Hence 1 and 4 are the correct matches.
B.
a = 2 $m/s^2$

$v_f=v-at$

putting $v_f$ = 0 to know how long will it take for block to reverse the direction.
0 = 4 - 2t
hance, t = 2 s

In 2 s, block will move a distance
$s=vt-\frac{1}{2}a{t}^2$
$⟹s=4\left(2\right)-\frac{1}{2}\left(2\right){\left(2\right)}^2$
$⟹s=4m$
Hence, it will not fall off the left end, instead it will reverse direction in between. Now, to know that when sliding stops, let's put $v_f$ = u,
-1 = 4 - 2t

$⟹t=\frac{5}{2}s$
displacement of object in t = 2.5 s will be,
$s=4\left(\frac{5}{2}\right)-\left(\frac{1}{2}\right)\left(2\right){\left(\frac{5}{2}\right)}^2$

$⟹s=\frac{15}{4}m$
The object displaces $\frac{15}{4}$ min 2.5 s while sliding. Now it will move $\frac{15}{4}$ m to the right at a constant speed that is 1m/s. So the time taken will be 15/4 s.
Total time = $\frac{5}{2}+\frac{15}{4}=\frac{25}{4}$ s
So the correct matches are 2, 3 and 5.

C.
a = 4 $m/s^2$

To know when the object velocity will be 0, writing equation,
0 = 2 - 4t

$⟹t=\frac{1}{2}s$
distance travelled in 0.5 s will be,
$s=2\left(\frac{1}{2}\right)-\left(\frac{1}{2}\right)\left(4\right){\left(\frac{1}{2}\right)}^2$
$⟹s=\frac{1}{2}m$
Hence, the block does not fall off the left edge, instead changes the direction in between. Now let's see if the sliding stops. To know this, let's equate v and u,
-5 = 2 - 4t
$⟹t=\frac{7}{4}s$
Displacement in this time,
$s=2\left(\frac{7}{4}\right)-\left(\frac{1}{2}\right)\left(4\right){\left(\frac{7}{4}\right)}^2$
$⟹s=\frac{-21}{8}m$
The value is negative that means the block would have fallen off right edge before sliding stops.
To know about the time, let's put s = 0,
$0=2\left(t\right)-\left(\frac{1}{2}\right)\left(4\right){\left(t\right)}^2$

$⟹t=1s$
The correct matches are hence 2 and 4.
D.
Since there is no frictional force, there is no acceleration.
$a=0.$
The box covers $10$ min $10$ s at a constant speed of $1m/s$ and falls off the left edge.
The correct match is $1.$