# [Energy] Work on a vertically thrown object

• brbrett
In summary, the conversation discusses a problem where a person throws a rock into the air and the question asks for the height the rock will reach, taking into account air resistance. The solution involves calculating the work done by the person throwing the rock, taking into account the energy lost due to air resistance. The correct answer is determined by subtracting the energy lost from the total work done, and this gives the potential energy at the top of the rock's flight, which can be used to calculate the height reached.

## Homework Statement

A person does 115 J of work in throwing a 0.325 kg rock straight up into the air. How high will the rock get, assuming there is 20 J of air resistance?

## Homework Equations

From what I formulated: W = ΔPE
The answer says E1 = E2, but I think that's the exact same as the formula I used.

## The Attempt at a Solution

W = ΔPE
115= mgh - 0
115 = (0.325)(9.8)h - 20
h = 42.386

It looked right in my head, but the problem is that the answer differ in the placement of the -20 J.
This is the correct answer given to me:
115 - 20 = Ep2
95 = mgh
95 = (0.325)(9.8)h
h = 29.83
30 J (Significant figured)

I don't understand why the -20 J should be placed with the work being done (the left side of the equation). I appreciate any attempt at helping me understand the reason.

The air resistance to which they refer is that which occurs between the instant the stone leaves the person's hand and the top of its flight. So the person has imparted 115J of energy to the rock, of which 20J is lost through air resistance on the upward journey, and the remainder is converted to potential energy.

Another way to look at it is that the air resistance pushes downwards on the stone, while the person, while throwing, was pushing upwards. So the work done by the air must have opposite sign to that done by the person.

• CWatters and Chestermiller
Alright, I think I understand. Thanks for the help!