The movement of a package on a conveyor belt

  • Thread starter Losmonkeys17
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Losmonkeys17
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Homework Statement
A package is dropped on a horizontal conveyor belt. The mass of the package is m = 0.563 kg, the speed of the conveyor belt is v = 0.947 m/s, and the coefficient of kinetic friction between the package and the belt is μk = 0.253.

a) How long does it take for the package to stop sliding on the belt?
0.382 s
You are correct.

b) What is the package’s displacement during this time?
0.181 m
You are correct.

c) What is the energy dissipated by friction?
0.252 J
You are correct.

d) What is the total work done by the conveyor belt?
Relevant Equations
W=-1/2*m*vi^2, W = Fd - fmgd
I have tried doing W=-1/2*m*vi^2 and total work = W + friction work
 
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  • #2
Losmonkeys17 said:
W=-1/2*m*vi^2
In that equation, how exactly is W defined? And doesn’t vi represent initial velocity?
 
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In what reference frame are you computing the work done? In what reference frame does the task want the work done to be computed?
 

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