Question

4. 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 m s-2, is
[AIPMT (Mains)-2011]
(2) 0.4 m
(1) Zero
(3) 1.2 m
(4) 0.6 m​

Answer 1

Answer:

(4)0.6

Explanation:

please vote me this answer is right

Answer 2

Acceleration = 0.5 x 10 (g = 10)
Using the third equation of motion
0 - 4 = 2 x 5 x s
S = -4/10 = 0.4 m relative to the belt.
Feel free to correct me!