Elastic Potential Energy and spring

[wallace13]

A 2.40 kg object is hanging from the end of a vertical spring. The spring constant is 40.0 N/m. The object is pulled 0.200 m downward and released from rest. Complete the table below by calculating the translational kinetic energy, the gravitational potential energy, the elastic potential energy, and the total mechanical energy E for each of the vertical positions indicated. The vertical positions h indicate distances above the point of release, where h = 0


I have already found kinetic and gravitational potential energy. I need to find elastic potential energy at h=0, .2, and .4m

Elastic Potential Energy= .5kx squared


.5(40)(.2)squared= .8 J

.8 J is the incorrect answer, but i know .5kx squared is the right equation... what am I doing wrong??


At this height (.2 above the release point) I found the correct kinetic energy to be .81 J and the correct gPE to be 4.7088 J

 

[Delphi51]

I'm confused over the point of release and the equilibrium point.
Looks to me like the point of release is h = 0, x = -0.2.
So at h = 0.2, x = 0 and the Elastic PE = 0.

 

[nlightner]

.

To find the elastic potential energy, you need to use the equation .5kx², where k is the spring constant and x is the displacement from the equilibrium position. In this case, the equilibrium position is at h=0, so x=.2m. Plugging in the values, we get:

Elastic Potential Energy = .5(40)(.2)² = .8 J

So, the elastic potential energy at h=0.2m is also .8 J. You may have made a calculation error or used the wrong values for k or x in your previous calculation. Double check your values and equations to make sure they are correct.