Diving board - energies

In summary, there is a question about determining the elastic potential energy of a 63 kg diver standing on a diving board that has compressed 8.0 cm. The book's solution involves considering the diving board as a spring and using the net force equation. However, some students argue that the gravitational potential energy should be equal to the elastic potential energy stored in the board. This is due to the fact that the system is not initially at equilibrium and some energy is lost due to the mass accelerating downward. The correct method involves providing a supplemental upward force initially and gradually backing off as the diver moves down, resulting in a balance of forces at the equilibrium position. Ultimately, the elastic energy stored in the spring is the same in both methods, but
  • #1
2
0
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?
 
Physics news on Phys.org
  • #2
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:
[tex]\int{Fdx}+ \frac{1}{2}kx^2=mgx[/tex]

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
 
  • #3
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.
 
Last edited:
  • #4
Thank you Chestermiller and Simon for clarifying the situation :)
 
  • #5


I would say that both methods are technically correct, but they are looking at the situation from different perspectives.

The first method considers the diving board as a spring and calculates the elastic potential energy based on the compression of the board. This method assumes that all the energy is being stored in the diving board and does not take into account the gravitational potential energy of the diver.

The second method takes into account the gravitational potential energy of the diver and assumes that this energy is equal to the elastic potential energy stored in the diving board. This method takes into account the energy of the entire system, including the diver.

The reason these two methods give different answers is because they are looking at the situation from different perspectives. The first method only considers the energy stored in the diving board, while the second method considers the energy of the entire system.

So, to answer the question of why the second method is incorrect, it is not necessarily incorrect, but it is not taking into account the full picture. It is important to consider the energy of the entire system when analyzing a situation like this.
 

What is a diving board?

A diving board is a long, narrow board that is typically made of fiberglass or wood and is used for diving into a swimming pool or other body of water. It is usually mounted on a spring or fulcrum, which allows it to bend and flex when weight is applied to it.

What are the energies involved in diving off a diving board?

The main energies involved in diving off a diving board are potential energy, kinetic energy, and the energy of impact. Potential energy is the energy stored in the diver as they climb the ladder or stairs to the diving board. Kinetic energy is the energy of motion as the diver jumps off the board and into the air. The energy of impact is the force that is exerted on the water when the diver enters it.

How does the height of the diving board affect the energy of impact?

The height of the diving board directly affects the energy of impact. As the height of the diving board increases, so does the potential energy of the diver. This results in a greater amount of kinetic energy and a more powerful impact with the water.

What safety precautions should be taken when using a diving board?

When using a diving board, it is important to follow all safety precautions, including checking the board for any damage or wear and tear, making sure the water depth is appropriate for the height of the board, and always diving in a straight line. It is also important to never dive onto a diving board from the side, as this can cause serious injury.

Are there any other factors that can affect the energies involved in diving off a diving board?

Yes, there are several other factors that can affect the energies involved in diving off a diving board. These include the weight and speed of the diver, the angle at which they enter the water, and any external forces such as wind or current. It is important for divers to be aware of these factors and adjust their diving technique accordingly to ensure a safe and successful dive.

Suggested for: Diving board - energies

Replies
1
Views
1K
Replies
1
Views
895
Replies
2
Views
644
Replies
3
Views
619
Replies
13
Views
898
Back
Top