A 2 kg package is released on a 53.1° incline, 4 m from a long spring with force

Question

A 2 kg package is released on a 53.1° incline, 4 m from a long spring with force constant k = 140 N/m that is attached at the bottom of the incline (Fig. 7-32). The coefficients of friction between the package and the incline are µs = 0.4 and µk= 0.2. The mass of the spring is negligible. (a) What is the speed of the package just before it reaches the spring? 7.299 Correct: Your answer is correct. m/s (b) What is the maximum compression of the spring? 0.97 Correct: Your answer is correct. m (c) The package rebounds back up the incline. How close does it get to its initial position? 0.736 Incorrect: Your answer is incorrect. m (short of the initial point)

Explanation / Answer

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...
=7.30 m/s

b)Now the spring compresses
energy stored in a spring
=.5*k*x^2
k=140 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(5...

this is a quadratic
a=.5*k
b=-(sin(53.1)*m*9.81-.2*cos(53.1)*m*9....
c=-.5*m*v^2
x= 0.97 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(...
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...
=.5*k*1.06^2-m*1.06*9.81*(sin(53.1)+.2...

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

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