Potential Energy in a vertical spring mass system

[karokr94]

Homework Statement

A mass of 0.5 kg hangs motionless from a vertical spring whose length is 0.90 m and whose unstretched length is 0.45 m. Next the mass is pulled down to where the spring has a length of 1.15 m and given an initial speed upwards of 1.4 m/s. What is the maximum length of the spring during the motion that follows?

Homework Equations

L=length of unstretched spring; (0.45m)

d=length of spring in equilibrium; (0.9m)

x=stretch=(d-L); (0.45m)

mg=kx ---> k=mg/x; (10.89N/m)

Equilibrium position = y = (L+x); (0.9m)

Usi+Ki=Usf
1/2k(1.15-y)^2+1/2mv^2=1/2k(max stretch)^2

The Attempt at a Solution

What I did at first was measure the stretch from the equilibrium position (0.25m) to get Usi and I solved for (max stretch) and added the (max stretch) to the length of the spring in equilibrium. I found some sources online, however, that say I should measure the stretch from the unstretched length? How does that even make sense? I would look up the answer in an answer key but I don't have one!

 

[TSny]

Hello and welcome

Your method will work. But note that you are not including changes in gravitational potential energy as the mass moves up and down! Why then does your method work? Turns out that measuring the stretch of the spring form the equilibrium position of the mass on the spring (rather than from the unstretched position of the spring) when calculating the spring's PE exactly compensates for leaving out the gravitational PE! It's a good exercise to show that.

If you measure the stretch from the unstretched position when calculating the PE of the spring, then you should also include the gravitational PE in your calculations.

 

[karokr94]

Hello and welcome

Your method will work. But note that you are not including changes in gravitational potential energy as the mass moves up and down! Why then does your method work? Turns out that measuring the stretch of the spring form the equilibrium position of the mass on the spring (rather than from the unstretched position of the spring) when calculating the spring's PE exactly compensates for leaving out the gravitational PE! It's a good exercise to show that.

If you measure the stretch from the unstretched position when calculating the PE of the spring, then you should also include the gravitational PE in your calculations.

That's what I thought but everything I could find online has me calculating the PE from the unstretched position without taking into account gravitational potential energy. I guess all the sources I found were wrong then! Kudos