# Gravitation Potential Energy Problem pls help

1. Sep 25, 2011

### starburst

Gravitation Potential Energy Problem pls help!!!

I would greatly appreciate help on a couple questions I have. I have been sic for the past few days and unable to go in to school to ask my TA's questions. My first question is when should use the equation Ki + Ui = Kf + Uf or Kf -Ki = Uf - Ui ? I know you would use both equations when there is conservation of energy but when would I use either one, aren't they pretty much the same?

And my second question is in reference to homework problem

1. The problem statement, all variables and given/known data

A rolelr coaster car (mass=988 kg including passengers) is about to roll down a track. The diameter of the circular loop is 20.0 m and the car starts from rest 40.0 m above the lowest point of the track. Ignore friction and air resistance. At what speed does the car reach the top of the loop?

2. Relevant equations

The solution the professor gave was change in K = change in -U --> Kf - Ki = -mg(hf-hi) --> 1/2mv^2 - 0 = mg(hi-hf) --> v = sqrt 2*g*(hi-hf) --> v = sqrt 2*9.8*(40-20) = 19.8 m/s

So my question in reference to this problem is why did he switch hf-hi to hi-hf, was that to get rid of the negative sign in front of mg? And if so how does that make sense? And also why did he set up the equation to be Kf -Ki = Uf - Ui instead of Ki + Ui = Kf + Uf?

Thank you very much i really appreciate it!

2. Sep 25, 2011

### dynamicsolo

Re: Gravitation Potential Energy Problem pls help!!!

One is just an algebraic rearrangement of the other. Writing it this way, Ki + Ui = Kf + Uf , says that the mechanical energy, E = K + U , is unchanged in the process of going from the initial situation to the final one. (This is true as long as friction, air resistance, transformations of work into internal heat, etc. aren't involved.) The other verson, which should read Kf - Ki = Ui - Uf = - ( Uf - Ui ) , means that the change in kinetic energy during the process is the opposite of the change in the potential energy ("when one goes up, the other goes down").

He is distributing the "minus sign" from the " - change in U" , as I did in the first comment above.

3. Sep 25, 2011

### starburst

Re: Gravitation Potential Energy Problem pls help!!!

Thank you so much for explaining that to me!

So why would they amount of energy at the top of the hill be the same as the amount of energy in the loop with a different height?

And are there specific types of problems/situations in which one would use one equation over the other? I believe i noticed that in problem when an object is just going from a height above tot he original height Ki + Ui = Kf + Uf was used. And in a problem where the object was going up and then back down Kf - Ki = Ui - Uf was used.

4. Sep 25, 2011

### dynamicsolo

Re: Gravitation Potential Energy Problem pls help!!!

Since we are treating friction and air resistance as unimportant (rather unrealistic, but it is easier to work out the problem for beginning physics students, since air resistance in particular is a more difficult force to deal with), the mechanical energy E = K + U can be treated as constant over the entire trip of the roller coaster. So we set

K (at start, where v is...?) + U (at start, h = 40 m) = K (we want to find this, so we can solve for vf at the top of the loop) + U (at top of loop, h = 20 m) .

There's no algebraic difference between the two versions of the equation. It can be that, depending on what information is given in the statement of the problem, that one version might be more convenient to use than the other. But using either version will give the same solution.

5. Sep 25, 2011

### starburst

Re: Gravitation Potential Energy Problem pls help!!!

Thank you very much for writing it out so clearly. Okay so the total conservation of energy in the system is the same. So if I were to solve the problem with the Ki + Ui = Kf + Uf method it would look like:

Ki + Ui = Kf + Uf --> 0 + mg(40) = 1/2mv^2 + mg920) --> mg(40)-mg(20) = 1/2mv^2 --> v = sqrt 2*m *mg(40)-mg(20)

Is this correct? Or could you possibly write out how the answer would look using Ki + Ui = Kf + Uf ? thank you

6. Sep 25, 2011

### dynamicsolo

Re: Gravitation Potential Energy Problem pls help!!!

Actually, the way we say this is that conservation of total mechanical energy means the total mechanical energy is the same throughout the process. (When we include "internal energy" of a system, we have "total energy", which is also conserved, which is the same as saying that it is constant during the process.)

Almost: notice that m , which is not zero, appears in every one of the terms. So it can be "divided out" across your next-to-last equation to give

mg(40) - mg(20) = (1/2)mv2 --> g(40) - g(20) = (1/2)v2

--> v = sqrt [ 2 * g * (40 - 20) ] .

Did you mean the other version with the changes in energy? That would be set up as

Kf - Ki = Ui - Uf

--> (1/2)mv2 - (1/2)m · 02 = mg(40) - mg(20)

[divide out 'm' ]

--> (1/2)v2 = g · (40 - 20)

and you would finish as you did in the earlier version.

7. Sep 25, 2011

### starburst

Re: Gravitation Potential Energy Problem pls help!!!

Oh I see I see, my algebra isn't the best lol. And also why is there a velocity at the top of the loop? I thought at the top of the loop there is greatest potential energy so no change in kinetic energy. Is it because the total potential energy is actually at a height of 40 m not 20 m?

There are actually two additional parts of the problem that I would appreciate helpi with if you can?

NEW QUESTIONS!

part (b) to the homework question is: What is the force exerted on the car by the track at the top of the loop?

Could I use the equation W = F*changeRcostheta and knowing that W = change in K set it so F*changeRcostheta = change in K ?

and part (c) From what minimum height above the bottom of the loop can the car be released so that it does not lose contact with the track at the top of the loop?

8. Sep 25, 2011

### dynamicsolo

Re: Gravitation Potential Energy Problem pls help!!!

The starting point is the top of the ramp of the roller coaster, which is at h = 40 m and where the car is at rest. So U = mg(40) and K = 0 there. The "ending point" (where we are comparing the situation) is the top of a loop in the ride: that point is at h = 20 m , and where we are calculating K , and vf from that result.

You won't get this part from the energy equation; we need to analyze the forces on the car at this point using a force equation.

The weight force mg on the car is pointing straight down at all times. The normal force N that the tracks apply to the car is also straight down at that instant. So the net force on the car is - ( N + mg ) . This is the physical force that produces the centripetal force that keeps the car moving on the circular loop at that point ; this resultant force points toward the center of the loop, which at that instant is straight down. So we can write

- ( N + mg ) = - ( Fc ) .

If you insert the expression for Fc , which depends on v and the radius of the loop, you'll see that we have everything needs to solve for the value of N.

The car will hold the turn on the loop, rather than fall off the loop before reaching the top, as long as N > 0 at the top of the loop. The absolute limit that can be tolerated is if N became zero only at the very top of the loop; it would be N > 0 for the car just before and just afterward. So we could set N = 0 in the force equation and figure out what v at the top of the loop would be in that circumstance (v there would be larger if N is positive there). We then back-figure using the conservation of energy equation: if v is the value we found with N = 0 at the top of the loop, what starting height on the roller coaster would the car have been at where v was zero when the ride began?