Conservation of energy - spring

[dba]

Homework Statement

A spring has an equilibrium length of 1m. The spring is compressed to a length of .5m and a mass of 2kg is placed at its free end on a frictionless slope which makes an angle of 41degree wrt the horizontal. The spring is then releases.
a. If the mass is NOT attached to the spring how far up the slope will the mass move before comming to rest.

Homework Equations

I thought since kinetic, restoring and gravitational energy is involved I start out with
1/2 mv_i^2 + mgy_1 + 1/2 kx_i^2 = 1/2 mv_f^2 + mgy_f + 1/2 kx_f^2

The Attempt at a Solution

Now I try to figure out what is actually active in my system.I first look at my
initial situation:
The spring is at rest, so v_i = 0 for my kinetic energy
The gravitational energy is present, since the block is present. My y_1 I need to calculate with y_1= 0.5m*sin41=.33m
I have potential energy of the spring also with x_i=-0.5m

Final situation:Here I do not know what acts in the system...
I have Kinetic energy with v_f
the potential energy of the spring is zero because x_f=0
But what to do with the gravitational energy? It says the block leaves the spring but is it still there in my final situation or not? Do I have to add it with the second hight y_f?


Then Part two when I found v_f of part one, I look again at
initial situation:
kinetic energy with v_i=v_f of part one
gravitational energy at y_i which I can set to zero

Final situation:
kinetic energy is zero, because the block comes to rest
gravitational energy at y_f which I need to find to find the distance.

Can someone please help me with this problem?
Thanks

 

[PhanthomJay]

What is part 2? In part 1, the final velocity is implicity given...solve for the height You can choose the initial height as 0 and then the final height is refernced to that 0 point.

 

[dba]

I wanted to split the problem in two pats.
part one: the spring travels until the mass got released.
part two: The mass leaves the spring and travels until rest.

I do not understand if I need the gravitational energy on the right side of my equation. And if I do, what would be the difference to the situation when the mass would be attached.

 

[PhanthomJay]

You don't have to split it into 2 parts, but it's OK to do so if you wish, at the risk of making errors due to the additional calculations involved. Yes, you do need the final Gravitational Potentiall Energy on the right hand side, since it surely exists with the 0 GPE point chosen at the start of the motion. If the mass is attached, both the elastic (spring) and gravitational final energies will be different than the unattached case, and you'll get a different final height.

 

[dba]

Ok, thank you. What I do not understand how I can set the gravitational potential energy zero at the start. It is mgy_1 and y_1 is not zero since the spring is not on the base. Can i just set a different base for the gravitational energy than for my spring potential energy?

 

[PhanthomJay]

Ok, thank you. What I do not understand how I can set the gravitational potential energy zero at the start. It is mgy_1 and y_1 is not zero since the spring is not on the base. Can i just set a different base for the gravitational energy than for my spring potential energy?

Yes, it's the change in gravitational potential energy that's important, so you can set the 0 point anywhere and get the same overall results.