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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 is µ=0.5. The distance that the box will move relative to belt before coming to rest on it, (taking g=10 ms2), is :

A
zero
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B
0.6 m
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C
1.2 m
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D
0.4 m
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Solution

The correct option is D 0.4 m
Given: μ=0.5,u=2m/s


When box will come in contact,

F=ma=μN

ma=μmg

a=μg=0.5×10

a=5m/s2

From the equation of motion,

v2=u2+2as
0=22+2×(5)s
s=25 w.r.t. belt
or distance =0.4m

Hence, (D) is the correct answer.

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