Gravitational Potential Energy, Elastic Potential Energy, and Kinetic Energy

[PeachBanana]

Homework Statement

A 4.0 kg mass is pressed down on a vertical spring of spring constant 400 N/m, compressing it to 0.250 m. After it is released, the amount of kinetic energy this mass would have when it leaves the spring is ___.

Homework Equations

mgy(final) + 1/2 kx^2 (final) + 1/2 mv^2 (final) = mgy (initial) + 1/2kx^2 (initial) + 1/2 mv^2 (initial)

The Attempt at a Solution

Ok so I think what I'm solving for is 1/2 mv^2 (final)

(4.0 kg)(9.8m/s^2)(0m) + 1/2 (400 N/m)(0m)^2 + 1/2(4.0 kg)(v)^2 = (4.0 kg)(9.8 m/s^2)(-0.250 m) + 1/2(400 N/m)(-0.250m)^2 + 1/2(4.0 kg)(0 m/s)^2

That was sort of long so to simplify it a bit:

1/2(4.0 kg)(v)^2 = (4.0 kg)(9.8m/s^2)(-.250m) + 1/2(400 N/m)(-0.250m)^2 + 1/2(4.0 kg)(0 m/s)^2

My main concern: My "x" and "y" are the same. Is that because it's a vertical spring? I'm calling the end of the release x=0 m and y=0 m and the compression -0.250 m.

 

[BruceW]

you have done it all correctly. Technically speaking, x is the displacement from the equilibrium point. So you can think of it as Δx = x₂-x₁, where the x₁ is the equilibrium point. We also know that the displacement Δx is purely vertical, so what is the (nice and simple) relationship between Δx and Δh?

 

[PeachBanana]

Ok, so Δx and Δh in this case are the same; that makes sense.

 

[BruceW]

yep, that's right. Often you'll find that its the change in height that is important in questions with uniform gravity.

 

[asteriatic]

I would like to clarify that the equations and attempt at a solution provided in this response are not entirely accurate. While the calculations for gravitational potential energy and elastic potential energy are correct, the equation for kinetic energy should be 1/2 mv^2 (final) = 1/2 mv^2 (initial) + 1/2 kx^2 (initial) - 1/2 kx^2 (final). This is because the spring is releasing the mass, so the initial velocity is zero and the final velocity is unknown. Additionally, the value for "x" should be negative since the spring is being compressed, not stretched. The correct equation and calculation for the final kinetic energy would be:

1/2 (4.0 kg)(v)^2 = 1/2 (4.0 kg)(0 m/s)^2 + 1/2 (400 N/m)(-0.250 m)^2 - 1/2 (400 N/m)(0 m)^2

= 0.03125 J

Therefore, the amount of kinetic energy the mass would have when it leaves the spring is 0.03125 J. It is important to use the correct equations and values in order to accurately solve for the final kinetic energy.