How Is Elastic Potential Energy Calculated in a Compressed Diving Board?

[dancerina]

This question came up in a textbook and not everyone agrees on the solution. The question is:

"A 63 kg diver is standing on a diving board waiting to commence her dive. The diving board has compressed 8.0 cm. Determine the elastic potential energy of the diver."

The solution given by the book is to consider the diving board as a spring and set the net force as zero, with the forces being that of the diver's weight and the spring's resistive force:

Fnet = mg - kx = 0

so k = (63 kg)(9.8 m/s^2)/(0.08 m) = 7717.5 kg/s^2

Then Ee = (1/2)kx^2 = (1/2)(7717.5 kg/s^2)(0.08m)^2 = 24.696 J

However, some students have said that the gravitational potential energy of the diver before they compress the board should be equal to the elastic potential energy that gets stored in the board; i.e.:

Eg = Ee, so mgh = Ee = (63 kg)(9.8 m/s^2)(0.08m) = 49.392 J (exactly double the previous answer).

They argue that the first method means 1/2 the Eg is lost somewhere (we're assuming this is an ideal situation). Can someone explain why this second method is incorrect?

 

[Chestermiller]

It all depends on how she got the the equilibrium position.

If she steps out on the board before the spring is compressed, then her weight is initially not supported by the board, and the system will not at equilibrium. She will begin to accelerate downward and develop KE at the same time that the spring is compressing. Her speed will continue to increase until she overshoots the equilibrium position, and then begins to slow down. But when she gets to the equilibrium position, her kinetic energy will account for the "lost energy" you referred to.

Method # 2 of reaching the equilibrium position is to provide a supplemental upward force equal to her weight initially, so that the system is initially in equilibrium. Then, you back on the force gradually as she moves down gradually, with the force being zero when she reaches the equilibirum position. So, during this process, your force balance becomes:

F -mg + kx = 0

If we integrate this equation from x = 0 to the equilibrium position x, we get:
\int{Fdx}+ \frac{1}{2}kx^2=mgx

So, the work done by the force F accounts for the "lost energy" you referred to.

In both cases, the elastic energy stored in the spring is the same.

Chet

 

[Simon Bridge]

The problem has the diver initially stationary on the board - we want to find the elastic PE, modelling the board as a spring. That should be the energy stored in the compression of the spring.

Should be true for any mass in equilibrium sitting stationary on top of a spring (or suspended from it) and should not depend on how the mass got there.

since$mg=kx$ then$$U=\frac{1}{2}\big(\frac{mg}{x}\big)x^2 = \frac{1}{2}mgx$$

Since x is the change in the height of the mass, then the mass has lost $mgx$ in gravitational potential energy.
If half of it is stored in the spring - what happened to the rest?

That about it?But it does matter how it got there!

When you get a mass on a spring, and you hold the mass at the springs unstretched length, and then you let go, the mass does drop to rest. Instead it bounces around. This is because some of the gPE went into kinetic energy. To come to rest, that kinetic energy has to go somewhere. i.e. into the spring...

But the mass won't naturally come to rest when there is zero force on it. When the force is zero, it keeps moving as before.

The position where the mass comes to rest is when $mgx=\frac{1}{2}kx^2$ ... at that point it is stationary all right, but it does not stay that way. The mass comes back up and oscillates around the position $x_0$ where the forces balance.

i.e. in general: $mgx_0 \neq \frac{1}{2}kx_0^2$

$mgx = \frac{1}{2}kx^2$ provided $|x-x_0|$ is the amplitude of the oscillations.

The diving board is a damped system ... the oscillations die away quickly.
The damping removes energy from the system.

 

[dancerina]

Thank you Chestermiller and Simon for clarifying the situation :)