Question

A block is gently placed on a conveyor belt moving horizontally with constant speed. After t=4s.
the velocity of the block becomes equal to the velocity of the belt. If the coefficient of friction
between the block and the belt is = 0.2. then the velocity of the conveyor belt is
(g= 10ms--
1) 2 ms-
2) 4 ms-
3) 64 ms -1
4) 8 ms-1​

Answer 1

Answer:

Option (4) 8 m/s

Explanation:

Acceleration of the block due to friction

$a=\mu\times g$

$a=0.2\times 10$

$a=2$ m/s²

Due to this acceleration the block acquires velocity v (velocity of the conveyor) starting from rest (gently placed, i.e. u=0)

From the first equation of motion

$v=u+at$

$v=0+2\times 8$

$v=8$ m/s

Therefore, the velocity of conveyor belt is 8 m/s

Hope this helps.