Package Dropped On A Conveyor Belt

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SUMMARY

The discussion focuses on a physics problem involving a package dropped on a conveyor belt, with specific parameters: mass of 4m, conveyor speed of 3v, and a coefficient of kinetic friction of 2k. The participant attempted to solve for the time taken for the package to stop sliding, its displacement, and the energy dissipated by friction. The calculations yielded symbolic results, including time as v/(26.16k), displacement as (17.44*(v^2))/(k), and energy dissipated as 1368.7*m*(v^2), indicating a misunderstanding of the need for numerical values in the final answers.

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Homework Statement


A package is dropped on a horizontal conveyor belt. The mass of the package is 4m , the speed of the conveyor belt is 3v , and the coefficient of kinetic friction between the package and the belt is 2k .

a) How long does it take for the package to stop sliding on the belt?

b) What is the package’s displacement during this time?

c) What is the energy dissipated by friction?

d) What is the work done by friction on the box?

Homework Equations



F = m*a
W = F*d

The Attempt at a Solution



I got all these problems wrong, and honestly I have no clue what I am doing with this problem. Here are my answers:

a) v/(26.16*k)
basically I calculated this by force of friction = (2k)(4m)(9.81)
then with F = ma
78.48mk = ma
a = 78.48k

So I used v = at
3v = 78.48kt
t = 3v/78.48k
t = v/26.16k

b) (17.44*(v^2))/(k)

basically I calculated this by force of friction = (2k)(4m)(9.81)
then with F = ma
78.48mk = ma
a = 78.48 k
Then d = .5at2
d = .5(78.48k)(v/26.16k)2

c) 1368.7*m*(v^2)

For this I used Ff*d

So Ff = 78.48mk
d = .5(78.48k)(v/26.16k)2

d) 1368.7*m*(v^2)

This is the same as c, right?
 
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It seems like all the results are going to be symbolic since you don't have numerical values for m,v,k. Why bother plugging in a numerical value for g? Why not just leave it as 'g'?
 

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