# Homework Help: Potential & Kinetic Energy

1. Dec 4, 2006

### CodeWebs

1. The problem statement, all variables and given/known data
A 2kg ball at the top of a 10m well has 0 potential and kinetic energy. The ball then falls to the bottom of the well and stops in the mud. After the ball stops what is the sum of its potential and kinetic energy?

2. Relevant equations
PE=m*g*h
KE=(1/2)M*V^2

3. The attempt at a solution
Ke + Pe = ?
(1/2)*2*0^2 + 2*(-10)*9.8 = -196 J
That is my answer, but my teacher says I'm wrong but cannot give me a reason.
Could someone please work out if I am right or what is the real answer and how you come to it.

Thank you

2. Dec 4, 2006

### dustpal

It seems right in the thinking but once you consider the conservation of energy, then it seems flawed. i believe the sum should still be 0...somehow..idk how else u could prove it cept with PEi + KEi = KEf + PEf >>>>>>0 = PEf + KEf

3. Dec 4, 2006

### CodeWebs

Well i assumed the energy of the system was still conserved because the energy of the ball was transfered to the mud. At any second before the ball hit the mud it would have a total of 0 J energy, but after its stopped it would then only still have the -196 J from its loss in potential. That was my thinking, it may be flawed.

4. Dec 4, 2006

### dustpal

well...mine was just a theory too, cuz we haven't covered PE and KE conservation but i just tried to remember it from last year.

5. Dec 5, 2006

### student85

hmm, it's strange to see a problem where the object has no PE when it's on top of it's final position.

If it had no PE then how did it have the potential to do work, that is to gain energy. Something mustve pushed it. I think the answer is 0 because all the energy is transferred to the mud.
I'm not sure though, wait for other replies.

6. Dec 5, 2006

### OlderDan

Assuming your teacher is correct, there is something about the problem that you are not taking into consideration. Did the ball penetrate into the mud some distance before stopping? Other than the possibility of some missing information, your calculation is correct.

7. Dec 5, 2006

### CodeWebs

The problem does not state any information about if the ball penetrated a certain distance into the mud, all it says is it stopped in the mud at the bottom. No matter how i show my teacher the calculations i did to find the answer, she says its wrong, yet she doesn't have any calculations to back up her answer. The other physics teacher in my school tryed to tell me that the reference point for the gravitational potential energy can be changed, and at the top of the well, the top is the reference point, and at the bottom its the bottom, yet from what i know this can't be done because it does not follow the law of conservation of energy. I feel as though I solved the problem as stated, and my teachers are only trying to back up the answer given to them rather than thinking about what is actualy taking place. I'm not sure what can be done if anything. If anyone has any further input into this problem it would be very helpfull, thank you.

8. Dec 5, 2006

### student85

This is just a question: Can energy really be negative? (PE or KE)

I too don't understand how the body can have 0 energy at first and have something later. There must be an external force that pushed the object, because it had no potential nor speed to do work.

9. Dec 5, 2006

### CodeWebs

I assumed the negative answer i got just meant that it would take 196 J to get back to its original position.

10. Dec 5, 2006

### OlderDan

I can't help you deal with your teachers, but I can tell you that the choice of the zero of gravitational potential energy is arbitrary. Once the choice has been made, as it was in this problem by stating the initial potential energy was zero at the top of the well, then you have to be consistent in that choice.

There is no reason why potential energy cannot be negative. In fact, when talking about planets and orbits the conventional choice is to define the zero of potential energy to be at infinity. When that choice is made, all gravitational potential energy is negative and given by the equation U = -GMm/r which is related to the gravitational force F = GMm/r²

You are correct about the conservation of energy. If the ball had zero potential energy at the bottom of the well, then its total mechanical energy would be zero, the same energy it started with, and that would mean that all of the energy that was imparted to the mud during the collision was created out of nothing. This would be a violation of conservation of energy. What happens to the 196J of energy that was lost by the ball? Some of it was converted into kinetic energy of flying mud that later became some potential energy of mud stuck to the sides of the well, but most of it was converted into heat energy of the ball and the mud that was dislodged in the collision.

11. Dec 7, 2006

### student85

OlderDan, one more question...
Your last post was really helpful btw, but then when we talk about an object having -176J of mechanical energy, I mean how can an object have negative energy? That´s my eternal question. Either it has or it doesn't..?

12. Dec 7, 2006

### OlderDan

Kinetic energy is positive and absolute in any given frame of reference. An object is either moving and has positive kinetic energy or it is at rest and has none. But even kinetic energy is relative. It depends on the velocity of an object relative to a specific frame of reference.

Potential energy is always related to a force that has a special property. We call such forces "conservative forces". The property is that if you move an object that is being acted upon by such a force, the net work done by the force is independent of the path taken by the object. Gravity is in this catagory. Potential energy is just a way of describing the work done by a conservative force when an object moves. The change in potential energy is equal to the work you must do to move an object against the action of a conservative force. It follows that the conservative force is the negative of the vector rate of change (negative gradient) of its related potential energy with respect to position. When the object moves to a position of lower potential energy, positive work is done by the conservative force.

The change of motion of an object is governed by forces (Newton's second law) and since the force only depends on the rate of change of potential energy with position, it does not matter what the value of the potential energy is at any one point. It only matters how fast it changes when you move from one point to another. If you graph gravitational potential energy (GPE) vs height near the surface of the earth, it is nearly a straight line, but only the slope of that line has any physical significance. There are infinitely many lines having the same slope, so you can choose any one of them to be the GPE.

On a larger scale, GPE is not a linear function of position. The force of gravity is -GMm/r² (- means attractive), which means I must do positive work to increase the separation of two objects, and that means that GPE must be -GMm/r + C where C is an arbitrary constant that has no effect on the slope of GPE. C has no physical significance, so we are free to choose any value we want for C. We could choose to call GPE zero at the surface of the earth, but if we did that we would have C = GMm/R where R is the radius of the earth and we would have to carry that constant term around with us every time we talked about GPE. Physicists prefer to be a bit more efficient and not carry around useless stuff, so the conventional choice is C = 0 and GPE = -GMm/r. Consequently, the GPE and the total energy of orbiting objects is always negative, but the kinetic energy is positive, and the GPE always increases as you increase r.

13. Dec 7, 2006

### student85

Thanks a lot OlderDan. You see energy has always been the topic I confuse the most. I can always do the problems and stuff, but just the concept of potential energy always bothered me.
I always thought of KE as the "real" energy and PE as the "fictitious" one, the one the object was CAPABLE of obtaining. Am I kind of right there or not?
I would really like your definition on energy btw, and thanks a lot in advance for all your help.

14. Dec 8, 2006

### fffff

error

At any instance when the ball is falling down, the energy at any particular point is constant.
K.E+P.E=E or constant

why do you say the ball has an initial potential energy of the ball is equal to zero. That is wrong. You cant have eithier P.E or K.E equal to zero. If your considering the potential energy of the ball from the bottom of the well, then its
(2)(9.8)(10)=196 J , K.E=0 because the ball has an initial velocity of 0 m/s.

The balls velocity when it makes contact with the mud
v^2=u^2+2gh
v&2=(2)(9.8)(10)
v=14 m/s

K.E of the ball as it makes contact with the mud=1/2(2)(14)^2=196 J
P.E at the bottom of the well
P.E = (2)(9.8)(0)

At any paricular instant the energy is a constant.

In your calculations, your taking the distance to be -10, which is totally wrong. Distance is a scalar, and so represents magnitude, not the direction. The same as with energy.

Your confusing yourself by taking the distance as -10m and the fact that P.E=K.E at the top of the well.

Last edited: Dec 8, 2006
15. Dec 8, 2006

### ZapperZ

Staff Emeritus
No, this is incorrect. You CAN have negative signs both for PE and "displacement", which is really a vector in the full definition. We just don't put the "i,j,k" when we're dealing with 1D motion.

Assigning PE to be zero at the top is perfectly valid. This is because PE is relative. What if the bottom of the 'well' is really on the 3rd floor of a building? Someone on the 1st floor can easily argue that his/her floor should be PE=0.

What should be considered is how PE changes - is the gravitational PE getting larger, or smaller, via that displacement. In this case, giving the displacement below the x=0 to be negative will make PE lower (i.e. more negative), and this is perfectly fine. If you don't trust me, try solving a simple problem such as dropping an object from a height h above the ground. Solve it both by using the ground as PE (and x) to be zero, and then translates the whole problem so that the starting point where the ball was drop to be PE (and x) to be zero. Let the ground be x = -h. You'll find the identical answer!

Note that it is crucial for you to be clear on this, because if you go on to do advance undergraduate mechanics and start dealing with Lagrangian and Hamiltonian, you will have to understand this concept very well.

Zz.

16. Dec 8, 2006

### fffff

can you give me another example please? I dont fully understand

17. Dec 8, 2006

### ZapperZ

Staff Emeritus
Which part don't you "fully understand"?

Zz.

18. Dec 8, 2006

### fffff

suppose if your dropping a ball from x metres
so its PE=mgx

PE at the ground =-mgx that is equal to KE at the ground
-mgx=1/2mv^2

how do you find the velocity if the sum is -ve?

19. Dec 8, 2006

### ZapperZ

Staff Emeritus
You are doing this TOO FAST. Write it down one step at a time.

What is the energy at the top? E(0) = U(0) + KE(0) = 0

What is the energy at the bottom?

E(-h) = U(-h) + KE(-h) = -mgh + 1/2 mv^2

Since we have conservation of energy, E(-h) = E(0) = 0

This means that

-mgh + 1/2 mv^2 = 0
1/2 mv^2 = mgh!

Zz.

20. Dec 8, 2006

### fffff

exactly what is U ?
if it is PE, why it is zero at the top?

Last edited: Dec 8, 2006
21. Dec 8, 2006

### ZapperZ

Staff Emeritus
Oy vey. You can't tell what it is when it has the form of "mgh"?

Zz.

22. Dec 8, 2006

### fffff

yes but why is it zero at the top. I though it was mgx

23. Dec 8, 2006

### cristo

Staff Emeritus
At the top the measurement of x is taken to be zero. Imagine drawing a diagram of this, a vertical line representing the wall, the top of which has coordinate 0, the bottom (the ground) has coordinate -h. Thus U(0) =mg.0=0

24. Dec 8, 2006

### ZapperZ

Staff Emeritus
<sigh>

I thought we were doing this the alternate way that I suggested EARLIER, in which one can set U=0 at the top where the mass started? Thus, since U(0) = mgx = 0 since x=0.

Then at the bottom of the well, since my "positive x" points up from the origin at the starting point, the bottom is at position x=-h. That's why U(-h) = -mgh.

Are we all clear now?

Again, you need to do this step by step and define things clearly from the very beginning. If not, the signs can kill you.

Zz.

25. Dec 8, 2006

### vanesch

Staff Emeritus
Your solution is entirely correct. However, in order to try to "educate" your teacher you might do it stepwise.

As you write, at that top, the ball is at rest, and has, according to the problem statement, also zero potential energy (this fixes the origin of the axis x in the formula U = m.g.x).
So we have E = U + T = m.g.0 + 1/2 m.0^2 = 0.

During its falling down, if we neglect air resistance, we have a conservative motion (meaning: E is conserved).
Hence, we have: E = m.g.x(t) + 1/2.m.(v(t))^2 = 0 (because it was 0 at the start, and remains conservedly 0).
This allows you to express the velocity as a function of the distance fallen (x, which must be negative, because the object is FALLING):
$$v = \sqrt{2.g.(-x)}$$ where (-x) is a positive number and so the square root has a meaning. At the bottom of the shaft, just before penetrating into the mud, the velocity is hence: $$v_{final} = \sqrt{2.g.(+10m)} = 14.0 m/s$$. If you plug this into the formula for the total energy, you will find of course that this is still 0, because that was what we required to find v in the first place.

However, enters the mud. The motion of the object in the mud is a thermodynamically dissipative process (you heat the mud in fact a little bit) in which mechanical work is converted in thermal energy. As such, conservation of energy restricted to the mechanical motion is not valid anymore. The object is loosing speed (while almost not changing its position and hence its potential energy). The total energy of the system will hence DECREASE, and fall to: $$E' = m.g.x$$ because v=0 at the end too. This is what you calculated correctly. It is negative, because (by convention) the initial energy was 0, and we lost some to the mud (which got heated up with exactly that amount of energy: 196 J). It is also negative, because you don't have enough mechanical energy anymore to get up to the initial position.

If we had choosen the origin of potential energy at the bottom of the well, then we would have 0 now, but we would have had an initial energy of +196J in the initial state.

Print this brilliant answer of mine out, and show it to your teacher

Last edited: Dec 8, 2006