Spring, Friction, Incline, Mass moving up a Ramp

[twilos]

Homework Statement

A 2.1 kg is released from rest from the bottom of a 45 degree inclined ramp. The package is attached to an ideal spring K = 35 N/m that is attached to the top of the incline causing the package to be launched up the incline. The coefficients of friction between the package and the surface are Us = 0.35 and Uk = 0.25

When the package is released, the spring is elongated 1.3m from its equilibrium position just before going up the ramp.

A. What is the speed of the package when the spring reaches its equilibrium position on the way up?

B. What will be the maximum compress of the spring?

Homework Equations

Kf + Ui = Ki + Ui + Other forces

Fs = 1/2kx^2

Ug = mg(sin45)

Us = 1/2kx^2

The Attempt at a Solution

I been trying to figure out how to set up the problem because i am unsure where Uspring and Ugravity are replacing in the conservation of energy. But here i go:

Kf + Ui = Ki + Ui + Other forces

1/2mv^2 + mg(sin45) = 0 + 1/2kx^2 - fkD

(.5)(2.1kg)(v^2) + (2.1kg)(9.8 m/s^2)(sin 45) = (.5)(35 N/m)(1.3m)^2 - (0.25)(2.1kg)(9.8 m/2^2)(cos45)(1.3)

also once you get the vf from part A you can find Part B's maximum compress which is K by using conservation of energy and substituting vf in and solving for K correct?

 

[twilos]

for Part B where you find the maximum compress I believe i am wrong and your actually finding X which is the distance the spring is compressed. Therefore when you find Vf you substitute it back into the conservation of energy equation and solve for X or D which i used both. And there will probably be some quadractic equation?

 

[LowlyPion]

Homework Statement

A 2.1 kg is released from rest from the bottom of a 45 degree inclined ramp. The package is attached to an ideal spring K = 35 N/m that is attached to the top of the incline causing the package to be launched up the incline. The coefficients of friction between the package and the surface are Us = 0.35 and Uk = 0.25

When the package is released, the spring is elongated 1.3m from its equilibrium position just before going up the ramp.

A. What is the speed of the package when the spring reaches its equilibrium position on the way up?

B. What will be the maximum compress of the spring?

Homework Equations

Kf + Ui = Ki + Ui + Other forces

Fs = 1/2kx^2

Ug = mg(sin45)

Us = 1/2kx^2

The Attempt at a Solution

I been trying to figure out how to set up the problem because i am unsure where Uspring and Ugravity are replacing in the conservation of energy. But here i go:

Kf + Ui = Ki + Ui + Other forces

1/2mv^2 + mg(sin45) = 0 + 1/2kx^2 - fkD

(.5)(2.1kg)(v^2) + (2.1kg)(9.8 m/s^2)(sin 45) = (.5)(35 N/m)(1.3m)^2 - (0.25)(2.1kg)(9.8 m/2^2)(cos45)(1.3)

also once you get the vf from part A you can find Part B's maximum compress which is K by using conservation of energy and substituting vf in and solving for K correct?

Welcome to PF.

Basically you are correct. Once you have the kinetic energy at the point that the spring is exerting no more force - changes from stretching to compression - then you use that kinetic energy to determine how much potential energy you can purchase and how much more friction will steal.

 

[twilos]

Welcome to PF.

Basically you are correct. Once you have the kinetic energy at the point that the spring is exerting no more force - changes from stretching to compression - then you use that kinetic energy to determine how much potential energy you can purchase and how much more friction will steal.

Just to clarify, that was all for part A right? Now to find the max compression of the spring do i keep all the variables still i.e kinetic, potential gravitational, potential spring, friction? And do i solve for k or x/D?

 

[LowlyPion]

Just to clarify, that was all for part A right? Now to find the max compression of the spring do i keep all the variables still i.e kinetic, potential gravitational, potential spring, friction? And do i solve for k or x/D?

I was referring to part B.

In A you have the PE of the spring being spent up until equilibrium on the PE increase, the KE increase and the expense of friction.

In B then it's the KE going back into Spring PE and gravity PE and the expense of friction.

As to solving, then of course it's x that you solve for since k is the same either way for the ideal spring.

 

[twilos]

Awesome i actually understood the concept now... =) so we're splitting Part A into the Before and Part B into the After. And Part B will be looked at again with the conservation of energy as "the KE going back into Spring PE and gravity PE and the expense of friction." correct?

 

[LowlyPion]

Awesome i actually understood the concept now... =) so we're splitting Part A into the Before and Part B into the After. And Part B will be looked at again with the conservation of energy as "the KE going back into Spring PE and gravity PE and the expense of friction." correct?

That's the idea.