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Q.

A package is at rest on a conveyor belt which is initially at rest. The belt is started and moves to the right for 1.38 with a constant acceleration of 2 m/s2. The belt then moves with a constant deceleration a2 and comes to a stop after a total displacement of 2.2 m. Knowing that the coefficients of friction between the package and the belt are μ =0.35 and μk=0.25, determine  the displacement of the package relative to the belt as the belt comes to a stop. (Take, g=10 m/s2 )

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a

0.33m

b

2.33m

c

5.33m

d

1.33m

answer is A.

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Detailed Solution

Acceleration of package will be 2 m/s2 while retardation will be μkg or 2.5 m/s2 not 6.63 m/s2.

For the package,

v=a1t1=2.6 m/ss1=12a1t12=1.69 m  

s1=vt2-12d2t22=2.6×0.4-12×2.5×(0.4)2  

=0.84 m  Displacement of package w.r.t. belt  =(0.84-0.51)m=0.33 m) 

=(0.84-0.51)m=0.33 m

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