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=10ms2), 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 D0.4 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