Package and Spring Physics?

Question

A 2.00-kg package is released on a 53.1 degree incline, 4.00 m from a long spring with force constant 120 N/m that is attached at the bottom of the incline. The coefficients of friction between the package and the incline are mu_s=0.40 and mu_k=0.20. The mass of the spring is negligible.

(a) What is the speed of the package just before it reaches the spring?

(b) What is the maximum compression of the spring?

(c) The package rebounds back up the incline. How close does it get to its initial position?

Answer 1

I like to use conservation of energy to solve these types of questions:

First, let's build a FBD of the package before reaching the spring

forces:
Parallel down-slope due to gravity
sin(53.1)*2*9.81
friction
.2*cos(53.1)*2*9.81

a) using conservation of energy for the 4.00 m slide
v=sqrt((sin(53.1)-.2*cos(53.1))*9.81*2*4)
=7.30 m/s

b)Now the spring compresses
energy stored in a spring
=.5*k*x^2
k=120 n/m
gravity continues to add energy to the ssytem as the package slides lower, and friction continues to burn energy as well
.5*k*x^2
=.5*m*v^2+x*(sin(53.1)*m*9.81-.2*cos(53.1)*m*9.81)

this is a quadratic
a=.5*k
b=-(sin(53.1)*m*9.81-.2*cos(53.1)*m*9.81)
c=-.5*m*v^2
x=1.06 m

Now the package is a total of 5.06 m downslope

on the return trip, the spring will decompress, friction and graivity will work against the upward sliding of the package

while the spring is decompressing the 1 .06 m
.5*m*v^2=.5*k*1.06^2-m*1.06*9.81*(sin(53.1)+.2*cos(53.1))
then the package will slide up the slope losing energy to potential energy and friction until it comes to rest in a distance d from the end of the spring. Since we want to know how far from the original position, calculate 4-d. I will just compute d
m*d*9.81*(sin(53.1)+.2*cos(53.1))=.5*m*v^2
=.5*k*1.06^2-m*1.06*9.81*(sin(53.1)+.2*cos(53.1))

d=2.68 m
so 4-d
=1.32 m from the start

j