# Package Dropped On A Conveyor Belt

## 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?

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?