July 4th and an example of work-energy

July 4th can be fun. One activity my family enjoys is playing in the lake at my parents house. Along with this comes the jumping off the dock. Great fun, and great physics. Here is a short clip.

Work Energy Example from Rhett Allain on Vimeo.

Notice that I violated my own rules for making videos. In particular, the camera was not perpendicular to the motion. Also, I can handle panning cameras, but not when there is nothing but sky in the background. This video is therefore not appropriate for a video analysis. That is ok. I don't need it to talk about physics. So, here is the question:

What is the average force exerted on the jumper when she is slowing down in the water?

To answer this question, you obviously need to think about forces. However, there are two main ways to think about force. The momentum principle:

And the work-energy principle:

The key difference is that the momentum principle deals with force and time where the work-energy principle deals with force and displacement. So, in this case it seems that we should use the work-energy principle since I can estimate how far the jumper falls and how far the water exerts a force on her. If I was looking at a rocket engine that fired for a given time, then I would use the momentum principle.

In this example what is the energy and what is the work? Really, this depends on what the system is. Picking the system is always the first step. I will choose the jumper plus the Earth as the system. This way, the only external force on the system would be the from the water. There would be kinetic energy of the jumper (the Earth essentially does not move) and gravitational potential energy of the jumper-Earth system. Since the jumper is close to the Earth, the gravitational field is mostly constant. Picture time.

I only represented one force from the water and not the gravitational force. Why? The reason is that I am going to use a gravitational potential energy, and thus there will not be a work done by gravity. You can't have work done by gravity AND gravitational potential energy. This would be like having your cake and eating your cake too. Can't be done.

Near the surface of the Earth, I can approximate the gravitational potential energy as only depending on height. Since I am really only interested in the change in potential energy, I can let y = 0 meters be where ever I choose. I choose the surface of the water to be y = 0 meters.

What about kinetic energy? This is easy. With the path I have chosen, the initial and the final kinetic energy are both going to be zero.

What about the work done? Only the force from the water does any work and this is only while the jumper is slowing down in the water. During this time, the force from the water is decreasing the energy of the jumper, thus it is a negative work. You can also see this because the displacement is down and the force is up making the angle between force and displacement 180 degrees. One way of writing the work is:

Where F and ?r are both the magnitudes of the two vectors (here is a vector review) and ? is the angle between these two vectors. So, putting this all together, I get:

Now, I can solve for the force:

And maybe I would like to get the acceleration - that way I can compare this to the NASA g-tolerance tables. To get the acceleration, I need to consider the net force on the jumper during this time. Here is a free body diagram:

So, putting in for the forces: (note that when I leave off the vector symbol, I am really referring to components of force. In this case, I am looking at the vertical components.)

Let me make some checks here. Will the acceleration be positive? Yes. The first term will always be positive and greater than g because (d+h)/d is greater than 1. What if a jumper jumps from a higher height (h). This would make the acceleration greater. What if the jumper stopped in less water, this would also make the acceleration greater. Finally, does this have the correct units? Yes.

Now for some estimations. Let me say that h = 3 meters and d = 1.5 meter. In this case, the acceleration is 19.6 m/s² or 2 g's. Referring to NASA g-tolerance table:

This is from wikipedia's page on g-tolerance - that table is no longer there. According to NASA, this would be in the +Gz position. You could handle 3.5 g's for up to 30 seconds. So, we are good.