# What is speed?

1. Oct 25, 2006

### Pizall

Hi, new at Physics so go easy.
My question is:
A roller coaster is traveling on a horizontal track at 20 m/s. What is its speed after climbing a hill 15m high? Ignore friction.

any help will be great.

2. Oct 25, 2006

Use the fact that the work done to lift a mass on a height h, equals the change of kinetic energy of that mass.

3. Oct 25, 2006

### Pizall

so it should be 1/2mv2 = 1/2mv1 - mgh
1/2v2 = 1/v1 - gh
v = square root of v1 - 2gh ?
Yes or No?

4. Oct 25, 2006

No, since this is the square root of a negative number. The work done is mgh, and the change of kinetic energy is 1/2 m(v2)^2 - 1/2 m(v1)^2, where v2 is the speed at the height of 15m, and v1 the initial speed. So, this implies 2gh + (v1)^2 = (v2)^2.

5. Oct 25, 2006

### Pizall

so it come up to v2 = 26.34 m/s

how is going faster going up the hill then what it started out?

6. Oct 25, 2006

I apologize - I overlooked something. The work mgh must be negative.

7. Oct 25, 2006

### JSBeckton

Its easier to think of this as consevation of energy-it initially has an energy of 1/2 mv^2, pure KE. Then it gains PE but for conservation to hold, it must lose KE

PE1+KE1=KE2+PE2

but PE1=0 because its at the base height of theis scenario.

KE1=KE2+PE2

1/2mv^2=1/2mv^2+mgh

the mass cacels

solve for v2

Last edited: Oct 25, 2006
8. Oct 25, 2006

Of course, it is clear what you wanted to say, but your equation 1/2mv^2=1/2mv^2+mgh implies 0 = mgh, and there is no 'v2' in it. So, be just a bit more precise when writing something.

Further on, it is not easier to think about this as conservation of energy, it's just another point of view.

9. Oct 25, 2006

### JSBeckton

I think conservation of energy

PE1+KE1=PE2+KE2

Clearly indicated that there were 2 KE's and therefore 2 velocities. And try not to tell people to be a bit more precise after you have just made a much more careless error. Mine was technical, yours was a mistake.

Furthermore, in introductory physics these concepts are developed through the conservation of energy law-always. I think that any physics teacher would agree that its necessary to understand the laws before trying to link work and PE.

I wouldn't be surprised if the particular chapter he is currently in was titled, "Conservation of Energy"

10. Oct 25, 2006

### Pizall

this is true it chp 4 4.7 Conservation of energy

so PE2 should be negative due to gravity

so KE1 = KE2 - PE2

11. Oct 25, 2006

### JSBeckton

Think about it this way, the initial energy must be equal to the final energy becasue energy cannot be created or destroyed, it can only change form. This is whats happening here, initially it has KE but no PE (PE is relevent to some reference point, we are choosing the initial reference point to be 0 for simplicity).

Now since it has gone up 15m from the reference 0 it has gianed PE, but where did this PE come from? It cannot be created so it must have come from a change, some of the KE transformed to PE.

Therefore the sum of energy to begin with must equal the sum of energy at the end.

the initial energy

PE+KE1, but at the begining you have no PE to the reference because we have chosen 0 to be the reference (0-0=0)

so the initial energy becomes

KE1

and that must be equal to the final energy which is:

KE2+PE2

so mathamaically it will look like this

KE1=PE2+KE2
initial=final

solve for V in the KE2 term

12. Oct 25, 2006

### JSBeckton

PE is only negative when something goes down from the reference point. It lhas lost PE.

13. Oct 25, 2006

### Pizall

so when I plug in the number I come up with a speed greater then What you started with. Started with 20 and after the math its 26 how is this

14. Oct 26, 2006

### JSBeckton

V2=sqrt(-2mgh+MV1^2)
=sqrt(-2gh+V2^2)
=sqrt[(-2)(9.81)(15)+20^2)]
=10.28 m/s

My guess is that you are STILL using

KE1=PE2-KE2

it should be

KE1=PE+KE2

you must be subtracting the 400 (the 20^2 term)

In which case you end up with a negative number (-694), should be a hint that you have made a mistake because you cannot take the sqrt of a negative number w/o complex numbers. Your algebra MUST be solid before you can tackle physics, no matter how many calculus/physics/engineering classes you ever take, you will find the without solid algebra, you will always struggle.

Everyone makes mistakes, its recognizing them that makes the difference, often the math will tell you that you need to go back and check something like in this case.

Hope this helps