Laws Of Motion Question 325

Question: A conveyor belt is moving at a constant speed of 2m/s. A box is gently dropped on it. The coefficient of friction between them isμ=0.5 . The distance that the box will move relative to belt before coming to rest on it taking g=10ms2 , is

Options:

A) 1.2 m

B) 0.6 m

C) zero

D) 0.4 m

Show Answer

Answer:

Correct Answer: D

Solution:

[d] Frictional force on the box f= μmg

 Acceleration in the box a = μg =5ms2

v2=u2+2aS0=22+2×(5)ss=25

w.r.t. belt  distance = 0.4m



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