Question

2. A conveyor belt is moving at a constant speed of
2 m/s. A box 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 coming to rest on it, taking g = 10 ms-a, is
[AIPMT (Mains)-2011]
(1) Zero
(2) 0.4m
(3) 1.2m
(4) 0.6 m​

Answer 1

Answer:

Explanation:

=> Here, it is given that

u = 2 m/s

μ = 0.5

g = 10 m/s²

=> Thus, retardation on the belt can be written as:

a = F/m

but, F = μmg

Thus, a = μmg / m = μg

=> Now, v²=u²+2as

0=2²−2*(μg)*s

0 = 4 - 2*(μg)*s

4 = 2*(μg)*s

∴ s = 4 / 2*μ*g

= 4 / 2*0.5*10

= 4 / 10

= 0.4 m

Thus, the  distance that the box will move relative to belt  before coming to rest on it is 0.4 m.