Problem with simple physics formula

[Artheas]

Ok, so i was thinking about thing to do with losing weight and energy usage, when i thought about the formula E(k) = (0.5) x m x v x v (half m v squared)

Lets just take m = 2 to simply things, making the formula E = v squared.

Taking the set distance as 100metres, if v = 1m/s, E = 1 joule per second, covering 1 metre per second, thus requiring 100J to cover 100m.

However if v = 2m/s, E = 4 Joules per second, covering 2 metres per second. This will take 50 seconds to reach 100m, so the total energy needed will be 50 x 4 = 200J ie double that at 1m/s.

This sort of made sense at first, since higher speed results in higher resistance (which is why c cannot be reached if i remember correctly), but for the energy requirement to double seems a bit strange to me.

Also, Gavin Hastings told me something along the lines of walking a mile uses the same energy as running a mile.

So is there a flaw in my calculations?

 

[geuben]

A half m v squared is the kenetic energy on object has, you want "work done = force x distance moved (in the direction of the force)"

 

[Visage]

Ok, so i was thinking about thing to do with losing weight and energy usage, when i thought about the formula E(k) = (0.5) x m x v x v (half m v squared)

Lets just take m = 2 to simply things, making the formula E = v squared.

Taking the set distance as 100metres, if v = 1m/s, E = 1 joule per second, covering 1 metre per second, thus requiring 100J to cover 100m.

No, E = 1 joule. Energy is measured in Joules. Power is measured in Joules per second, or Watts. Inthe absense of resistive force no extra energy is required to travbel 100m. For example, in space, you can set an object in motion and it will continue forever, with no need for any extra energy.

However if v = 2m/s, E = 4 Joules per second, covering 2 metres per second. This will take 50 seconds to reach 100m, so the total energy needed will be 50 x 4 = 200J ie double that at 1m/s.

Again, incorrect, for the same reason above.

This sort of made sense at first, since higher speed results in higher resistance (which is why c cannot be reached if i remember correctly), but for the energy requirement to double seems a bit strange to me.

No, it does not make sense. You've taken an equation that has no input in terms of the magnitude of resistive force and somehow you've managed to come up with an answer that magically takes into account resistance - doesnt tat strike you as being somewhat odd?

Also, Gavin Hastings told me something along the lines of walking a mile uses the same energy as running a mile.

Its possible, I guess - you'd use up less energy per second walking, but you'd obviously be doing it for a longer time.

So is there a flaw in my calculations?

Yes, many.

 

[Bill Door]

The formula you have used will give the amount of kinetic energy that is posessed by the runner (or object) which is not necessarily the same as the amount of energy required to sustain that velocity (unless you are running in treacle).

I'm not really sure how you could accurately calculate the energy requirements of walking a mile compared to running it, since both expend large proportions of energy in avoiding falling compared to energy spent propelling forward. An example of this would be to consider how much energy it would take to cycle a mile (on a flat gradient) at a given speed compared to running it.

 

[Artheas]

Geuben and Bill answered it perfectly for me, thanks for the replies.

I was confused since E = mgh does allow you to calculate the energy used when climbing a certain height. I just thought of the two formula as being similar in many ways when we were taught them many years ago.

E = mgh is essentially the same as "work done = force x distance moved (in the direction of the force)"(in a vertical direction), whereas E = 1/2 m v squared evidently is not

 

[Van_Dammesque]

Geuben and Bill answered it perfectly for me, thanks for the replies.

I was confused since E = mgh does allow you to calculate the energy used when climbing a certain height. I just thought of the two formula as being similar in many ways when we were taught them many years ago.

E = mgh is essentially the same as "work done = force x distance moved (in the direction of the force)"(in a vertical direction), whereas E = 1/2 m v squared evidently is not

The 1st equation you stated was kinetic energy (half m, v squared), the last (m ,g ,h) is the potential energy (done on earth).

@ Visage, Im a physicist too, experimental, you seem like a theorist if my gut instincts tell me correctly!

 

[Visage]

@ Visage, Im a physicist too, experimental, you seem like a theorist if my gut instincts tell me correctly!

My undergrad degree had a large chunk of experimental physics, including my final year project, but my postgrad work was more theoretical, being very methematical in basis.

 

[daz]

I'm a final year undergrad physicist too.

 

[Sleepy]

Don't you think you 3 brain boxes should have picked up the error in this then?

I was confused since E = mgh does allow you to calculate the energy used when climbing a certain height. I just thought of the two formula as being similar in many ways when we were taught them many years ago.

E = mgh is essentially the same as "work done = force x distance moved (in the direction of the force)"(in a vertical direction), whereas E = 1/2 m v squared evidently is not

 

[Psyk]

Also, Gavin Hastings told me something along the lines of walking a mile uses the same energy as running a mile.

The same amount of energy has been used to move your body over that distance, but since our bodies are less efficient the faster we move (generally speaking) lots more energy will have been wasted if you run, mostly in heat form. That's why it actually uses up much more of your energy if you run, even though the same amount of energy has actually been used to get you from a to b.

 

[Visage]

The same amount of energy has been used to move your body over that distance

Im not convinced that thats true - resistive forces obviously increase if you run....

 

[Dave]

whereas E = 1/2 m v squared evidently is not

Work is in terms of Nm, or (kg m s^-2)(m) or kg m^2 s^2. Those are the units given by mv^2, so it evidently is the same unit representation

 

[Psyk]

Im not convinced that thats true - resistive forces obviously increase if you run....

Well I suppose that's true, but in a sense the same amount of energy has been put into moving that distance, but also more has been used to overcome the resistive forces. If any of that makes sense.

 

[Visage]

Well I suppose that's true, but in a sense the same amount of energy has been put into moving that distance, but also more has been used to overcome the resistive forces. If any of that makes sense.

Well, in the absense of resistance it doesnt take any energy to move from a to b. If you fire a bullet from point a to point b then it has the same energy at b that it had at a.

With resistance, however you need to continually put energy in to counteract resistive forces. Put simply, power = F * v where F = resistive force and v = velocity

 

[Psyk]

Well, in the absense of resistance it doesnt take any energy to move from a to b. If you fire a bullet from point a to point b then it has the same energy at b that it had at a.

With resistance, however you need to continually put energy in to counteract resistive forces. Put simply, power = F * v where F = resistive force and v = velocity

Yeah I suppose you're right. I suppose I was thinking more about moving against mavity.

 

[Artheas]

Don't you think you 3 brain boxes should have picked up the error in this then?

I'm not sure whether you mean the first or second part? (or both)

If it's the first part, your E(potential) is at x. So after you go up h, the new potential energy you have is y. so y-x should be roughly what you expended to get there.

If it's the second part, if energy used = force x distance. vertically, F=ma (mg)
So E = mgh is an 'implementation' of work done = force times distance

... right? =/

Work is in terms of Nm, or (kg m s^-2)(m) or kg m^2 s^2. Those are the units given by mv^2, so it evidently is the same unit representation

That sounds like you're saying half m v squared = work done? the suggests a joke but i'm getting confused now.

Please excuse my lesser brain.

 

[Sleepy]

Well, in the absense of resistance it doesnt take any energy to move from a to b. If you fire a bullet from point a to point b then it has the same energy at b that it had at a.

With resistance, however you need to continually put energy in to counteract resistive forces. Put simply, power = F * v where F = resistive force and v = velocity

In this case I suggest that F is a function of m, t, v and v^2

 

[Sleepy]

I'm not sure whether you mean the first or second part? (or both)

Energy 'used' to climb a hill is not a simple case of KE => PE work is done in overcoming friction, drag, and within the body to create the force needed to move it etc

 

[Visage]

In this case I suggest that F is a function of m, t, v and v^2

Yes, thats entirely possible.