### Video Transcript

A woman walks along a bridge, holding in her hand a small cell phone with a weight of 0.75 newtons. The phone is held at a height of 1.2 meters above the surface of the bridge. The bridge is 4.3 meters above the surface of a river. How much greater is the gravitational potential energy of the phone held in the woman’s hand on the bridge than the gravitational potential energy of the phone at the surface of the river?

Okay. To start out, let’s clear some space to make a sketch of this situation. Let’s imagine that this is the woman holding this cell phone while she’s walking along a bridge. And we’re told that the height of the cell phone above the surface of the bridge is 1.2 meters. And we’re further told that the bridge itself is above a river. And that the distance between the bridge deck and the river surface is 4.3 meters. We want to contrast the gravitational potential energy of the phone when, in one case, it’s in the woman’s hand and, in another case, when the phone is down at the surface of the river. Maybe she’s dropped it, unfortunately.

To get started on our solution, let’s recall the mathematical relationship for the gravitational potential energy of some object. We can abbreviate that GPE. And it’s equal to the mass of an object multiplied by the acceleration due to gravity multiplied by the height of that object above some reference level. In our case, that reference level is the surface of the river. We can call that a height value of zero meters. That’s our standard or our baseline. And like we said, what we want to solve for here is the difference in the gravitational potential energy of the phone when it’s in the woman’s hand compared to when it’s on the surface of the water.

We could call this difference ΔGPE. And we know that’ll be equal to the gravitational potential energy of the phone when it’s held in the woman’s hand minus the gravitational potential energy of the phone when it’s on the river. Here’s how we’ll write that out. We’ll say that ΔGPE, that change in gravitational potential energy, is equal to the gravitational potential energy of the phone when it’s at the top, that’s when it’s in the woman’s hand, minus the gravitational potential energy of the phone when it’s at the bottom, when it’s on the surface of the water.

Now that we have this equation, we can apply our general relationship for the gravitational potential energy of a mass of object. So then, ΔGPE is equal to the phone mass times the acceleration due to gravity times the height of the phone at the top of its position, when it’s in the women’s hand, minus the phone’s mass times lowercase 𝑔 times the height of the phone when it’s just on the surface of the water. Looking at the right-hand side of this equation, we see that the factors of 𝑚 and 𝑔 are common to both terms. That means we can factor them out. So, this change in gravitational potential energy is equal to the phone mass times 𝑔 times the change in its height, ℎ sub 𝑡 minus ℎ sub 𝑏. Now that we’ve written out these heights, let’s label them on our sketch.

We know that ℎ sub 𝑡 is the height of the phone when it’s being held in the woman’s hand. That’s up here. And we know that ℎ sub 𝑏 is the height of the phone when it’s just above the surface of the water. That’s right here. But now notice something about ℎ sub 𝑏. That’s equal to where we’ve set our zero reference for height. We’ve said that ℎ sub 𝑏 is equal to ℎ, which is equal to zero meters. So, if we go to our equation for the change in gravitational potential energy, we can cross out ℎ sub 𝑏 and replace it with a zero. Which means that our overall expression reduces to 𝑚, the mass of the phone, times 𝑔, the acceleration due to gravity, multiplied by ℎ sub 𝑡, the height of the phone when it’s in the woman’s hand. And what is that height? Well, it’s the sum of the height from the river to the bridge and from the bridge to the woman’s hand. 4.3 meters plus 1.2 meters. And if we add these two terms together, we get a result of 5.5 meters.

So, now all that remains is for us to plug in for 𝑚 and 𝑔, the mass of the phone and the acceleration due to gravity. And actually, we don’t have to plug in for them separately. But we can do it all at once. And the reason for that is, we were told in the problem statement that the weight of the phone is equal to 0.75 newtons. Well, we’ll write that this way. We’ll say capital 𝑊 is equal to 0.75 newtons. But what is the weight of a mass of object? Well, we know the weight of a mass of object, capital 𝑊, is equal to 𝑚 times the acceleration of the gravitational field it’s in. In our case, since we’re on the surface of the Earth, that acceleration is lowercase 𝑔, 9.8 meters per second squared. This all means that we can replace 𝑚 times 𝑔 in our expression with our value for 𝑊. That is, we can replace it with 0.75 newtons.

All that’s left for us to do now is to multiply this value by the distance, 5.5 meters. When we do, we find a result of 4.125 joules. That’s the difference in the gravitational potential energy of the phone from when it’s in the woman’s hand to when it’s on the surface of the river.