The movement of a package on a conveyor belt

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The discussion focuses on the calculation of work done on a package moving on a conveyor belt, specifically using the equation W = -1/2*m*vi^2. Participants clarify that W represents the work done, while vi denotes the initial velocity of the package. There are questions about the reference frame used for computing work, highlighting the importance of context in physics problems. The conversation emphasizes the need to specify the reference frame for accurate calculations. Understanding these concepts is crucial for correctly applying the work-energy principle in this scenario.
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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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Losmonkeys17 said:
W=-1/2*m*vi^2
In that equation, how exactly is W defined? And doesn’t vi represent initial velocity?
 
In what reference frame are you computing the work done? In what reference frame does the task want the work done to be computed?
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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