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Question

A package is kept on a conveyor belt and the system is at rest. The belt starts to move to the right for 1.3 s with a constant acceleration of 2 m/s2. The belt then moves with constant deceleration a m/s2 and comes to a stop after a total displacement of 2.2 m. Knowing that the coefficient of static friction between the package and the belt is 0.35 and coefficient of kinetic friction is 0.25. Then determine the displacement of the package relative to the belt as the belt comes to stop. Take g=10 m/s2.



A
0.33 m
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B
0.66 m
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C
0.22 m
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D
0.11 m
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Solution

The correct option is A 0.33 m
The FBD of the package when the belt is moving with constant acceleration a1=2 m/s2


Let mass of the package be m
The package will be at rest relative to the belt as external force on it is ma1=2m and the limiting frictional force is fl=μsN=0.35mg=3.5m

Now, the velocity of the package/belt is
v=a1t1=2×1.3=2.6 m/s
Displacement of the package/belt, when it moves with constant acceleration is
s1=12a1t21=12×2×(1.3)2=1.69 m

Given, total displacement is 2.2 m
Now, the displacement of the belt during retardation is
s2=(2.21.69)=0.51 m
Also, from the equation of motion
s2=v22a2
 a2=v22s2=(2.6)22×0.51=6.63 m/s2
and t2=va2=2.66.63=0.4 s

Relative motion will start only when the belt starts to decelerate.
So, the friction force acting on the package will be kinetic in nature, which is given by
fk=μkN=0.25mg=2.5m

Let the package acceleration relative to the belt be ar.
Hence, writing the force equation


ma2fk=mar
 6.63m2.5m=mar
 ar=(6.632.5)=4.13 m/s2

Thus, relative displacement is given by
Sr=urt2+12art22=0+12×4.13×(0.4)2
=0.33 m

Physics

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