### Video Transcript

How much work is done against the gravitational force on a five-point-zero-zero-kilogram briefcase when it is carried from the ground floor to the roof of the Empire State Building, a vertical climb of 380 meters?

In this problem, weβll assume that the 380-meter height is an exact number, and weβll also assume that π, the acceleration due to gravity, does not change over this vertical extent. Weβll treat it as a constant and exact value of nine point eight meters per second squared.

A simple sketch of this scenario will help orient us to be able to solve this problem of figuring out the work done against the gravitational force. Now here, we have a picture of our Empire State Building, and not drawn to scale is our briefcase which is a five-point-zero-zero-kilogram mass that we lift up to the roof of the building a height weβre told is 380 meters.

Now we know that in order to raise the briefcase up this height, weβll have to work against gravity which tends to push down on the briefcase through the weight force. So we want to know what is that work done against gravity; will call it π sub π for work against that must be done in order to raise the briefcase through this height.

In this example, the work done by gravity is negative because gravity pushes down on the briefcase, yet the briefcase moves up. The work done against gravity, π sub π, is positive.

And in fact, in terms of the magnitudes of these two work values, the magnitude of the work against gravity is equal to the magnitude of the work of gravity. So what is the work that gravity does on this briefcase?

To figure this out, we can recall the fact that the work done by gravity in this case is equal to negative the change in potential energy of the briefcase. So if we solve for the potential energy of the briefcase at the bottom or at the ground level β weβll call it PE sub π β and the potential energy of the briefcase at the top of its position β weβll call it PE sub π‘ β that change will equal negative the work done by gravity. So what is the potential energy of the briefcase at the bottom and at the top of its position?

Letβs define height to be zero at the base of the Empire State Building, where the briefcase starts out. And letβs also recall that the gravitational potential energy of a mass π is equal to that mass times π, the acceleration due to gravity, times the height of the object.

Now looking at our sketch, since weβve defined height to be zero at the base of the tower, we know that PE sub π, the potential energy of the briefcase as it starts is zero and PE sub π‘, the potential energy at the top of the briefcases position, is equal to the mass of the Briefcase times π times 380 meters.

Multiplying those three numbers together, we get the result of 18.6000 newton-meters or 18.6 kilojoules.

Now that weβve solved for the potential energy of the briefcase at its highest position, we can find the change in potential energy, π₯PE, by subtracting the potential energy at the bottom from the potential energy at the top.

The overall change in potential energy of the briefcase is 18.6 kilojoules minus PE sub π, or minus zero.

So if that is the overall change in potential energy of our briefcase β we said earlier that the work done by gravity is negative that change. So in other words, the work done by gravity is equal to negative 18.6 kilojoules.

And we can work back from that to solve for π sub π, the work done against the gravitational force on this briefcase. Weβve established that the magnitude of π sub π is equal to the magnitude of π sub π, and in fact theyβre equal and opposite. That means that π sub π, the work done against the gravitational force as this briefcase is lifted up, is equal to positive 18.6 kilojoules.

This work is positive because the force exerted to create this work as well as the displacement of the briefcase are in the same direction. Now this is just one of a few ways that one can solve this problem.

Another way could be to use the fact that work equals force times displacement. In that case, we could have used the mass of the briefcase along with π, the acceleration due to gravity, and π, the displacement of the briefcase, to solve for π sub π. Several different ways, but we end up with the same result.