- #1

disregardthat

Science Advisor

- 1,861

- 34

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter disregardthat
- Start date

- #1

disregardthat

Science Advisor

- 1,861

- 34

- #2

cristo

Staff Emeritus

Science Advisor

- 8,107

- 73

but we only got a change in position for the object.

What do you mean by this? Remember that acceleration is a vector quantity, and also that v=0 does not imply a=0, and so in your case you have a change in acceleration.

- #3

- 136

- 1

- #4

disregardthat

Science Advisor

- 1,861

- 34

haiha: So it must turn to some sort of energy, for example heat you say?

- #5

Doc Al

Mentor

- 45,051

- 1,364

Of course, if you add or remove KE, that energy must come (or go) from somewhere.

- #6

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 963

Think of it this way: compress a spring, doing work to store potential energy in the spring. Now use that spring to accelerate your object.

Quickly run around in front of the object, place your spring in front of it and use the spring to slow it. Ignoring resistance or any kind of friction, the kinetic energy in the object will be exactly enough to compress the spring back to where it was before. If you don't like the idea of "running around in front of the object", imagine two different but identical springs. Same thing happens. Net result, you did work to compress a spring initially, and now you are back in that same situation. No energy change.

- #7

- 905

- 4

The answer here is that the kinetic energy you put into the object was later transformed either into potential energy somewhere else, or into heat. For example, if you use a spring to stop the object, the kinetic energy from the formerly moving object will be transformed into spring kinetic energy. Or if you stop an object by friction (or some other non-conservative force), then the kinetic energy will be lost to friction. I could give slightly different answers based on the mechanism used to stop the object, but this is the basic idea.

- #8

disregardthat

Science Advisor

- 1,861

- 34

So it's actually just a wrong perspective of seeing it from the beginning. I see it now. Does momentum count as energy, or more particulary: Is momentum kinetic energy?You did work on the object when accelerating it. The object did work on you when decelerating it.

- #9

- 863

- 4

Yes and no. They are not the same thing. However, you may notice that momentum "p" is the derivative of kinetic energy with respect to velocity:

[tex]\[

\begin{array}{l}

E_k = {\textstyle{1 \over 2}}mv^2 \\

p = mv \\

\end{array}

\]

[/tex]

[tex]\[

\begin{array}{l}

E_k = {\textstyle{1 \over 2}}mv^2 \\

p = mv \\

\end{array}

\]

[/tex]

Last edited:

- #10

disregardthat

Science Advisor

- 1,861

- 34

I see, thanks.

- #11

russ_watters

Mentor

- 20,413

- 7,012

Also, if you give something (like a car) a push, then run around to the other side of it and stop it, you'll expend roughly the same amount of energy stopping it as you did in starting it moving, since your muscles aren't springs.

In Hurkyl's example with the springs, the end result looks exactly like a mirror image of the initial setup and you're back to zero as if nothing had happened. In an example with a person, you've converted a lot of chemical energy to heat.

And to meet the two in the middle, if you use the car's engines to proved the push then the brakes to provide the stopping, it works exactly the same from an energy standpoint as if all the engine's initial energy input had been turned directly to heat.

In Hurkyl's example with the springs, the end result looks exactly like a mirror image of the initial setup and you're back to zero as if nothing had happened. In an example with a person, you've converted a lot of chemical energy to heat.

And to meet the two in the middle, if you use the car's engines to proved the push then the brakes to provide the stopping, it works exactly the same from an energy standpoint as if all the engine's initial energy input had been turned directly to heat.

Last edited:

Share: