# Energy Conservation and Friction

akhila1489

## Homework Statement

A 2.00 kg is released on a 53.1degree incline, 4.0 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 static friction constant = 0.40, and kinetic friction constant= 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?

## The Attempt at a Solution

K1+U1+Wother = K2+U2
Wother= -fs
s is 4m, the distance it needs to get to the spring.

f= fk + fs
fk= 0.20n
fs=0.40n

how do i get n? (n=wcos53.1??)

in that case f= fs+fk= [0.20(2*9.8)(cos53.1)]+[0.40(2*9.8)(cos53.1)]=7.06
f= [-(K2+U2)-(K1+U1)]/s
f=[-(0+mgh)-(0.5mv^2 + 0)]/4 = 7.06
28.24=(-2*9.8*h)-(0.5*2*v^2)

to solve for v, we need h, is h simple sin53.1?

(*i made a diagram as well, but it isn't allowing me to post it...*)

Last edited:

akhila1489
i got part c by doing the following

let x = max compression

U1-fs=0.5kx^2
s= 4+x​
k=120N/m​
2*9.8*(4+x)sin53.1-2.35(4+x)=60x^2
solving the quadratic gives you x=1.06m.

is this remotely correct?

Homework Helper
Welcome to PF!

in that case f= fs+fk= [0.20(2*9.8)(cos53.1)]+[0.40(2*9.8)(cos53.1)]=7.06
f= [-(K2+U2)-(K1+U1)]/s
f=[-(0+mgh)-(0.5mv^2 + 0)]/4 = 7.06
28.24=(-2*9.8*h)-(0.5*2*v^2)

to solve for v, we need h, is h simple sin53.1?

Hi akhila1489! Welcome to PF!

Yes, h is 4sin53.1.

but I can't follow the rest of what you've done for part a).

Just use KE + PE = constant!

Homework Helper
Hi akhila1489,

## Homework Statement

A 2.00 kg is released on a 53.1degree incline, 4.0 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 static friction constant = 0.40, and kinetic friction constant= 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?

## The Attempt at a Solution

K1+U1+Wother = K2+U2
Wother= -fs
s is 4m, the distance it needs to get to the spring.

f= fk + fs

You don't want to add the kinetic and static frictional forces. You need to decide which one applies to the case of a object sliding down an incline plane, and use that.

fk= 0.20n
fs=0.40n

how do i get n? (n=wcos53.1??)

in that case f= fs+fk= [0.20(2*9.8)(cos53.1)]+[0.40(2*9.8)(cos53.1)]=7.06
f= [-(K2+U2)-(K1+U1)]/s

I don't think this follows from your original equation. Your original equation was

$$K_1+U_1+W = K_2+U_2$$

so this means that K2 and U2 are the final values, and K1 and U1 are the intial values. You also had W= -f s. So when you solve for f, I think you have a sign error; it should be:

$$f=[-(K_2 +U_2) + (K_1 + U_1)]/s$$

(Also, later you plug in final values for K1 and U1, and initial for K2 and U2.)