# Befuddled By Basics

1. May 26, 2005

### JO 753

I am having trouble making sense of the KE=1/2MV^2 formula.

If I am interpreting it correctly, it is stating that twice the speed = 4 times the energy.

To my way of thinking, it seems to violate conservation of energy and relativity. I even did a simple experiment ( dropped a weight on some springs ) but it only added to my confusion.

Not having any higher math training, what I need is some real world example that clearly demonstrates this formula.

2. May 26, 2005

Staff Emeritus
It doesn't violate conservation because you had to put in that much energy to get up to that speed. It takes four times as much energy to get up to 2V as it does to get up to V. So of course you then HAVE four times as much energy stored, as it were, in your speed.

3. May 27, 2005

### JO 753

Lets say you have a rocket in space. You set your speedometer at zero, then burn 100 kilos of fuel and now the speedometer says 1000 kph.

Are you saying it will take another 300 kilos of fuel to make that speedometer show 2000 kph?

4. Jun 1, 2005

### JO 753

Is this too basic for everybody here or do none of you have an answer?

5. Jun 4, 2005

### JO 753

Here iz the experiment I did:

I dropped a wate far enuff to compress 2 springs by 5mm. Then I added 6 more springs to the device and dropped the wate from a hite that woud cause it to be going twice az fast az the 1st drop. But because of the crude setup, I needed to drop it from more than 4X the hite in order to compress the 8 springs by 5mm. ( expected )

So here iz my quandary:

4X as many compressed springs = 4X az much energy.

4X the force for haf the time = twice the energy.

Both statements seem undeniably true. The experiment supports both statements, yet they contradict each other.

Wut am I missing? ( and dont say 'haf a brain'! )

6. Jun 4, 2005

### learningphysics

Yes, that's right.

7. Jun 4, 2005

### learningphysics

Yes, the above makes sense.

The above isn't true. Force times distance (not time) gives energy. How are you getting 4X the force? The weight is the same so the force acting on it is the same.

8. Jun 4, 2005

### JO 753

I am getting 4X the force because each compressed spring has the same amount of force and there are 4X az many of them.

The way I see it, if you used an geared down electric motor to compress the springs and its controlled to take the same amount of time to compress them, it will require 4 times the wattage to compress 4 times the springs.

But, if you double the speed, it will still need to supply the same amount of force, but for only haf the time, therefor haf the watt hours.

The falling wate will cover the 5mm compression distance in haf the time if its going twice az fast.

Is it not true that it takes only twice the energy to stop something in haf the time?

Or is the falling wate actually hitting with 8 times the force?

9. Jun 4, 2005

### learningphysics

Remember that the force the spring exerts is changing. F=kx where x=extension (or compression) of the spring.

4 identical springs would exert F=4kx.

Yes. Here I'm assuming that we are dealing with a horizontal spring and ignoring gravity.

You lost me here. Remember that the force the spring exerts changes with time. The power delivered by the spring to the mass is changing with time.

I don't think this is true. I don't think that in the 4 spring system the distance is covered in half the time. Remember that the acceleration isn't constant.

The same something? You have to do the same amount of work to stop the object both times (the work is negative).

10. Jun 5, 2005

### JO 753

The falling wate will compress the springz in haf the time if its going twice az fast. The average speed from wutever to zero iz haf the wutever speed. So twice the average speed over the same distance iz haf the time.

Maybe if i describe the device better, you can get a good idea of wut I'm thinking, learningphysics.

The device I made has a rod to guide the falling wate, which is a 1/2 pound brass cylinder. It lands on a plate that has pockets for 8 springs. It has some retainers to precompress the springs to a set amount & 2 latches which keep it from bouncing back up after the kenetic energy is expended. The distance from the preload to the latched position is 5mm. If you drop the wate from an insufficient hite for the number of springz, the plate will not latch. Dropping the wate too far iz eazy to see because it bounces. You can see that each spring will have a set average force over the 5mm of travel.

Obviously this device iz not made for any accurate mesure, since it duznt take the continuous force of gravity on the resting wate into account. It iz plenty accurate for the pupose of seeing the general amount of kenetic energy, tho. I knew befor I made it that, if KE=1/2MV^2 iz correct, then I woud need 8X az many springz to stop the wate when it wuz going twice az fast. It wuz not made to hold 16 springz. It works with 2, 4, 6, or 8 springz.

So, with the motor example, lets say you set your controller to make it take ten times az long to compress the springz. This will require it to supply that average amount of force for ( for example ) 50 seconds instead of 5. The force duz not decrease because its going slower, so the energy used will be 10 times az much.

11. Jun 14, 2005

### learningphysics

Hi JO. Sorry not to have replied for so long. A couple of points I wanted to make:

1) Energy delivered = (Average Power over time) * (total time)

So average power (over time not distance) is the average energy delivered per unit time.

Energy delivered = (Average Force over distance)* (total distance)

So average force (over distance not time) is the average energy delivered per unit distance.

So in the case of a motor compressing a spring... if you want the motor deliver the same amount of energy in half the time.... then the average power has to be double... but not necessarily the average force!

2) Let's look at a horizontal spring situation. If we shoot some object at some velocity v towards the spring... suppose it takes time t to get to velocity 0 from the time it hits the spring.

Now if we shoot the object at 2v... and we use 4 springs. The time the object takes to come to rest is 0.5t... half the time with one spring and velocity v.

The average power (over time) delivered to the springs 8 times as much as before.

Power factor * time factor = Energy factor

8 * (1/2) = 4

The average force (over distance) is 4 times as much as before. Distance is the same as before.

Force factor * distance factor = Energy factor

4 * 1 = 4

I've just summarized the working of the spring... But to actually derive the stuff takes a little math. I just wanted to show the force, power and time factors.

12. Jun 15, 2005

### Integral

Staff Emeritus
I have moved this to the General Physics Forum, It should get more exposure there.