Earlier, you read about a puzzle involving a ball falling to the Earth. We resolved this puzzle by introducing the concept of potential energy. In these notes, you will look an example of a person lifting a box using different choices of system. The goal here is to demonstrate that the choice of a system matters.

A person is lifting a box (of mass $m$) from the ground to a height $h$. At that height, the box is moving with an upward speed $v$. To lift the box, we assume the person applies a force $F$ upward. Let's consider three different choices of system to demonstrate what can be learned from each choice.

While lifting the box, the person's internal energy will change, so we will keep track of that change with $E_{person}$.

• System: Box, Earth, and Person; Surroundings: Nothing
• Initial State: Box on ground with no velocity; Box at height $h$ with velocity $v$

$$\Delta E_{sys} = W_{surr}$$ $$\Delta K_{box} + \Delta U_{grav} + \Delta E_{person} = 0$$ $$\dfrac{1}{2}mv^2 + mgh + \Delta E_{person} = 0$$ $$\Delta E_{person} = -\left(\dfrac{1}{2}mv^2 + mgh \right)$$

Because the system is only a single particle, there's can be no potential energy. There is work done by both the person and the Earth, but they have opposite signs because the applied force and the gravitational force are in different directions.

• System: Box; Surroundings: Earth and Person
• Initial State: Box on ground with no velocity; Box at height $h$ with velocity $v$

$$\Delta E_{sys} = W_{surr}$$ $$\Delta K_{box} = W_{person} + W_{Earth}$$ $$\dfrac{1}{2}mv^2 = Fh - mgh$$ $$\dfrac{1}{2}mv^2 = \left(F-mg\right)h$$

Finally, this system has potential energy, but there is work done by the person.

• System: Box and Earth; Surroundings: Person
• Initial State: Box on ground with no velocity; Box at height $h$ with velocity $v$

$$\Delta E_{sys} = W_{surr}$$ $$\Delta K_{box} + \Delta U_{grav} = W_{person}$$ $$\dfrac{1}{2}mv^2 + mgh = Fh$$