Question

5.
A conveyor belt is moving at a constant speed of 2
m/s. Abox is gently dropped on it. The coefficient
of friction between them is u = 0.5. The distance
that the box will move relative to belt before com-
ing to rest on it, taking g=10 ms?, is
[NEET - 2011]​

Answer 1

Answer:

0.4metre

Explanation:

let initial speed=i, which is equal to 2m/s.

let final speed=v,is equal to zero.

u=0.5.so force of friction between them is umg.now acceleration will be (umg/m) that is equal to ug.

so by Newton's law's of motion,v^2-i^2=2as.now by putting the formula we get,0^2-2^2=2*0.5*10s.

so s is equal to 4/10=0.4metre.

Hope it helps.