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Question

A block of mass 1 kg is kept on an inclined accelerating conveyor belt. Acceleration of the belt is 1 m/s2 as shown in figure. Minimum coefficient of static friction between the belt and block to prevent slipping of the block will be: [Take g=10 m/s2]


A
0.4
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B
0.3
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C
0.2
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D
0.69
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Solution

The correct option is D 0.69
m=1 kg
Acceleration of belt, a=1 m/s2
Considering μmin, limiting friction (f=μminN) should be sufficient to ensure that the block moves with same acceleration as the bellt, to avoid slipping.

FBD of block:


Applying equation of dynamics along direction of acceleration,
fmgsin30=ma
f=ma+mgsin30
f=(1×1)+1×10×12=6 N ...(i)
From equilibrium condition along perpendicular to the inclined surface,
N=mgcos30
N=1×10×32=53 N ...(ii)
Substituting for limiting friction (f),
f=μminN
6=μmin×53
i.e μmin=653=253=0.69

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