# Homework Help: Conservation of Energy Homework problem

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1. Oct 16, 2016

### jlmccart03

1. The problem statement, all variables and given/known data
A ball of mass m falls from height hi to height hf near the surface of the Earth. When the ball passes hf, it has a speed of vf. Ignore air resistance. Consider the system T which consists of the ball only.

Write an expression for each of the following quantities in terms of the given variables and any physical constants:
The net work on system T
The change in kinetic energy of system T
The change in potential energy of system T

2. Relevant equations
ΔKE = 1/2mv2
ΔPE = -mgh = -mg(hf - hi)
Wnet ext = ΔKE

3. The attempt at a solution
I know that the total energy throughout the system does not change. However, Potential Energy caused by gravity does change into Kinetic Energy as the ball falls for the initial height. Therefore I believe that the net work on system T is ΔKE = 1/2mvf2.

The change in kinetic energy is ΔKE = 1/2mvf2

And the change in Potential energy is 0.

This is where I felt that I may be completely wrong since the net work is all the forces multiplied by displacement I think my net work formula or "expression" is incorrect. Could anyone help me verify my mistake? Thanks!!!

2. Oct 16, 2016

### LemmeThink

Hi!

First of -
Could you list all the forces acting on the ball?

Now,
The expression you have used here is the Work-Energy theorem. What is the fundamental definition of work? What about the potential energy?

3. Oct 16, 2016

### Tallus Bryne

Do you see that your ΔKE expression implies an initial state of rest (if what you mean by v is vf) ? You did not state that as an assumption, so you might want to make sure that it's an assumption worth making.

Either the PE changes by some nonzero amount or it doesn't change. You can't have both.

4. Oct 16, 2016

### jlmccart03

Okay, the only force acting on the ball in this scenario is weight or mg.

The fundamental definition of work is W = F * d or force times displacement. Potential energy due to gravity (Since technically we are dealing with PEgrav) is defined as PE = m*g*h or mass times gravity times height.

So I am assuming that the work done on the system T would be mgΔh as indicated by the definition of work?

Last edited: Oct 16, 2016
5. Oct 16, 2016

### LemmeThink

That is correct.
No. Work done, would be, as you correctly stated, $\vec F.\vec d$, along with the appropriate sign. You should not be using PE for determining the work done. While you could make use of the potential energy, it is better to work with the definition $W = \vec F.\vec d$ So tell me, is gravity doing positive, or negative work?

6. Oct 16, 2016

### jlmccart03

I realized what I said originally was wrong and fixed it within my reply.
It is most definitely doing positive work since the motion of the ball and gravity are in the same direction correct?

7. Oct 16, 2016

### LemmeThink

Yes; and it's value?

8. Oct 16, 2016

### Tallus Bryne

If you consider the positive direction for the motion to be "up", and the ball is falling "down", then the work could be expressed as -mgΔh as long as we consider the
Δh to be negative due to the choice of orientation for our coordinate system.
However, like what LemmeThink said, you don't really need to consider potential energy in your approach here. The force of weight being -mg, and the displacement d = Δh gives you the work using the fundamental definition. From this expression for work we can come to our definition of ΔPE as the negative of the work done by gravity.

9. Oct 16, 2016

### jlmccart03

It would be mgΔh correct? Since the force is mg and d is the height h.

10. Oct 16, 2016

### jlmccart03

Oh, okay, I think I get it now. From the work expression I can then determine my energy formulas (or at least my potential energy formula) since the net Work is equal to these two formulas or am I still way off from the target?

11. Oct 16, 2016

### Tallus Bryne

We're dealing with gravity, but the following approach is supposed to work with any force that can be considered conservative (means that work done by that force is path independent). You derive an expression for the work done by that force over some specified path, which begins at point A and ends at point B, or initial and final heights, for example. The result can be used as a basis for defining a potential energy for that force, equating the negative of the work done by the force to the change in that potential energy.
So instead of using the work-kinetic energy theorem as a starting point in a problem, where the net work expression involves all the forces acting on the object, you can start with a statement of the conservation of energy: ΔKE + ΔPE = W , where W is now just the work done by all the non-conservative forces (like friction as a common example in intro physics problems), and the work expressions for all conservative forces are moved to the "energy side" of the equation.

12. Oct 16, 2016

### jlmccart03

Okay so, the net work done on the system is mgΔh, ΔKE = 1/2mvf2 and PE = -mgΔh in this scenario correct? This is based on the definition of work and definition of PE.

13. Oct 17, 2016

### Tallus Bryne

This is fine if you treat Δh as a positive number (this would work with a coordinate system with the downward direction being positive). Perhaps I'm being a little too pedantic about consistency with signs and coordinate systems. The most important thing is that your result for work is positive, ΔKE is equal in value but opposite in sign to ΔPE (no matter what your coordinate system). So in your case that makes ΔKE positive and ΔPE negative by the same amount.