Physics

# The Physics Philes, lesson 16: Appreciate the Gravity of the Situation

In which balls are thrown, gravity works, and energy is added.

Holy cow, you guys. This is the 16th post of this series and, frankly, I long ago ran out of clever ways to introduce each installment. (You may have noticed.) So I’m just going to jump right into it.

Last week I attempted to explain gravitational and elastic potential energy. I’m running a little behind this week, so I’m going to have to save elastic potential energy for next week. I spent too much time last week at the Renaissance Festival and watching the special features on my Blu-ray of The Avengers. But for now, it’s game on, gravitational potential energy. GAME. ON.

Let’s say that you throw a 0.145 kg ball straight up in the air. By throwing the ball you give it an upward velocity of 20.0 m/s. If we ignore air resistance (which, of course, we will do), how high will the ball fly?

Because the ball is being thrown straight up, we don’t need an x-axis. I’ll treat point one as the point at which the ball leaves your hand and point two where the ball reaches it’s maximum height. The equations we need are:

and

OK. Which variables do we know? As the free body diagram shows, our target variable is y2, the ball’s maximum height. Luckily, we know everything else:

m = 0.145 kg

v1 = 20.0 m/s

v2 = 0 m/s

g = 9.80 m/s

y1 = 0

The velocity at point two is zero because the ball will have reached its potential height and will instantaneously be at rest. Since y1 = 0, the potential energy at point one is zero. The same is true for kinetic energy at point two; since the ball is at rest, kinetic energy at point two is zero. Do you understand why this would be? Remember that potential energy is mass times gravity times position. If the position is at zero, then the whole equation must equal zero. Kinetic energy is 1/2 times mass times velocity. If the velocity is zero, the entire equation must equal zero. Dig?

So we know that potential gravitational energy at point one is zero and we know that kinetic energy at point tow is zero. From this we know that:

So we need to find K1. To find K1, we just need to input some variables:

This is equal to the gravitational potential energy at point two, which is mgy. (Remember, y2 is the target variable.) We just have to rearrange the equation to find y2:

Voila! Neglecting air resistance, that ball went 20.4 m in the air. But what if we have other forces doing work on the ball, as well? What if your hand moved up 0.50 m while you’re throwing the ball. Your hand once again gives the ball an upward velocity of 20.0 m/s. Assuming the force your hand exerts is constant, what is the magnitude of the force? And what is the speed of the ball 15.0 m above where it left your hand?

In this case, there is nongravitational work being done, so we need this equation:

In this situation, we have three points: y1 = 0.50 m; y2 = 0 m; and y3 = 15.0 m. We also know that the velocity at y1 = 0 m/s (because the ball is at rest) and at y2 = 20.0 m/s. The velocity at y3 is one of our target variables. The nongravitational force, which is our other target variable, only works between points y1 and y2. Let’s determine the nongravitational force first.

To figure out the nongravitational force, we need to calculate the other work indicated on the left side of the above equation. To do that, we need to find the other variables:

Potential gravity at point one is negative because the ball was initially thrown below the origin, but don’t worry about that. To find the other work done, we need to add the difference between the kinetic energy at points one and two and the difference between the gravitational potential energy at those same points:

So the other work done is equal to 29.7 J. Now how do we use this number to find the force? Remember the work equation? W = Fs, where s is the displacement. The displacement in this case is (y2 – y1) because the other force only acts in between those points. By rearranging the equation, we have:

Great! We’ve found the force! Now we need to find the speed at point three. Between points two and three, we don’t need to worry about other work, so we can just use the equations from the first problem.

Since K3 = 1/2(mv^2), where v is the y-component of the ball’s velocity at point three, we have:

Do you see that plus or minus sign? That’s because the ball will pass point three twice, once on the way up and once on the way down.

When I first encountered these equations, I was a little intimidated. Especially when they are expressed in symbols. For a while I was getting the kinetic energy at point one confused with kinetic energy with point two and the same with the potential energies. But as I spent some time with the equations, it all became clear. Or clearer, at least.

I wish I had time to get into elastic potential energy this week, but I can’t. Sorry. Next week we’ll explore elastic potential energy in all it’s snappy glory!