Why kinetic energy is proportional to velocity squared

[Emanresu]

Could someone please explain why kinetic energy is proportional to velocity squared and not just velocity.

A propulsion system that provides a constant force would typically use energy at a constant rate in order to accelerate at a constant rate and thus rate of change of velocity would be proportional to rate of use of energy. So why is rate of gain of energy not also proportional to rate of change of velocity ?

E.

 

[pixel01]

Consider an object is moving at v. If you exert a force F to change its speed dv in time dt and the distance dS. Then the work needed is :
A = FdS= ma.dS
There must be one v in a and another in dS.

 

[stewartcs]

Just to give a visual aid, a little more info can be found here...

http://hyperphysics.phy-astr.gsu.edu/hbase/ke.html

 

[mordechai9]

In short, kinetic energy is proportional to velocity squared, rather than velocity, or velocity cubed, or something else, for the following reason. Simply, the conservation of energy imposes this proportionality. The only real way to see this is to see the mathematical origin of the term "kinetic energy". The quantitative value of kinetic energy arises from quantitative (mathematical) consideration of the physical situation.

Basically, what you're looking for is a certain principle, the so-called "work energy theorem", or principle of conservation of energy. The derivation of this principle is extremely simple, and it will tell you EXACTLY why energy is proportional to velocity *squared*.

Let's consider the simplest possible case of a rigid body with a constant mass moving in 1-direction. The body starts from rest, at an arbitrary position in space. From Newton's discovery, we know:

1. F = ma = m \frac{dv}{dt}

Next, we want to consider what would happen if a force were applied to the rigid body as the rigid body moves from one point to another. So, we multiply each side by an infinitesimal increment of length, and integrate:

2. int(F ds) = int(m \frac{dv}{dt} ds)

Next, we realize that we can use a trick from calculus to rewrite the right hand side, considering:

3. \frac{dv}{dt} = \frac{dv}{ds} \frac{ds}{dt} = \frac{dv}{ds} v

Putting (3.) into (2.), we get:

4. int(F ds) = int(m v \frac{dv}{ds} ds) = int(m v dv)

Integrating the left hand side you get a value which represents force applied over a distance; this is precisely the quantity known as "work". Integrating the right hand side, you get a term which represents .5*m*v^2. This is precisely the quantity which is defined as "kinetic energy". Hence, the proportionality of KE to v^2 comes from the consideration that changes in kinetic energy should be directly proportional to the work done on an object.

I hope this helps. There's a lot of math and you may not understand everything. It seems like a long-winded answer but really this is the complete answer. I don't really understand the second part of your question; you ask, "why is rate of gain of energy not also proportional to rate of change of velocity ?", and yet, rate of gain of energy IS proportional to rate of change of velocity... it's just not linearly proportional.

 

[Emanresu]

Thanks for your responses, but I'm still a bit confused. I understand the math and how kinetic energy is calculated as work done (force x distance). It's just that it doesn't make sense to me when I consider a propulsion system. If you are putting energy in at a rate proportional to time (to create a constant force) how can you get energy out at a rate proportional to time squared (kinetic energy) ?

 

[jtbell]

If you are putting energy in at a rate proportional to time (to create a constant force)

A constant force on an accelerating object does not deliver energy to the object at a constant rate. The rate at which energy is delivered per unit time is P = F.v, which increases as v increases.

Or, if you deliver energy at a constant rate P, then the force must decrease as the velocity increases, and the acceleration decreases accordingly.

 

[pixel01]

Jtbell explains very clearly. I just add an example. A falling object (free falll) always has a constant force exerted on : the gravity. But as time elapses, the height reduces faster, that means the potential energy is converted into dynamic energy at higher rate.

 

[Emanresu]

A constant force on an accelerating object does not deliver energy to the object at a constant rate.

I don't doubt this is true, I just don't understand it. Why does it take more energy to accelerate from 5 to 10 m/s than from 0 to 5 m/s ? Are these not the same just in a different frame of reference ?

 

[mordechai9]

By my calculations considering a 1 kg object under gravitational force, the object obtains 12.5 Joules of energy accelerating from 0 to 5 m/s, versus 37.5 Joules to go from 5 to 10 m/s. So, the object actually acquires more energy accelerating from 5 to 10 m/s than from 0 to 5 m/s. The object doesn't "take energy" to move faster; I think you understand this incorrectly. The object takes force to move faster; and it becomes more energetic as a result. I guess you could say it takes more energy to move an object faster, the faster it is already going. However, it seems incorrect to say that it "takes more energy" to move an object faster, because objects do not feel energy directly; they only feel forces.

Now, as jtbell said, a constant force on an accelerating object does not deliver energy to the object at a constant rate. Why? You tell me! We don't make the rules of nature. We just figure them out. This is the universe man and you can only explain so far.

Maybe, intuitively, you can understand it by thinking that it takes more energy to make an object move faster, after it is already moving very fast. If you want to kick a soccer ball from rest, you can kick it pretty damn far. But if you want to kick a soccer ball that is already flying very fast, it is hard to kick it so that it moves a lot faster.

 

[Emanresu]

Let me try a different wording.

Say you have a rocket ship at rest with a certain amount of chemically stored energy (fuel). To provide a constant force to constantly accelerate the rocket the chemically stored energy is used up at a constant rate. This energy gets mainly turned into kinetic energy and heat.

Energy cannot be created or destroyed. The fuel energy is being used at a constant rate and producing heat at a constant rate but producing kinetic energy at an ever increasing rate. It appears that energy is being created out of nowhere !

I am missing something here, what is it ?

 

[jtbell]

The fuel energy is being used at a constant rate and producing heat at a constant rate but producing kinetic energy at an ever increasing rate.

If you burn fuel at a constant rate (kg/sec or liters/sec or whatever) and thereby produce energy at a constant rate (joules/sec), I guarantee that your rate of acceleration will not remain constant, but will decrease as time passes, so that the kinetic energy increases at a constant rate. Specifically, v as a function of time will be proportional to the square root of t.

 

[Emanresu]

Thanks for guaranteeing that JT, but I'd prefer if you would explain it. Otherwise I'll have to use the Chewbacca Defense.

E.

 

[ZapperZ]

mordechai9 has given you a detailed derivation of this. In what sense are you still confused? As far as I can tell, you are trying to hold on to an a priori requirement that the change in velocity must be a constant. There's nothing "fundamental" here with respect to this change in velocity. There is something fundamental with conservation of energy, and that should always be your starting point. And the derivation given already in this thread is based on such conservation.

Zz.

 

[TVP45]

I don't doubt this is true, I just don't understand it. Why does it take more energy to accelerate from 5 to 10 m/s than from 0 to 5 m/s ? Are these not the same just in a different frame of reference ?

The thing to remember is that the energy produced by a gallon of gas will be MEASURED differently in different frames.

 

[Emanresu]

The thing to remember is that the energy produced by a gallon of gas will be MEASURED differently in different frames.

That sounds promising. Assuming you understand why I am confused that energy is not proportional to v, can frame of reference be used to explain why it is proportional to v squared ?

E.

 

[Emanresu]

mordechai9 has given you a detailed derivation of this. In what sense are you still confused? As far as I can tell, you are trying to hold on to an a priori requirement that the change in velocity must be a constant. There's nothing "fundamental" here with respect to this change in velocity. There is something fundamental with conservation of energy, and that should always be your starting point. And the derivation given already in this thread is based on such conservation.

Zz.

That is why I gave an example where 'it appears to me' that conservation of energy is violated ...

 

[learningphysics]

I think the major concept to remember here is that force is not the time derivative of energy... power is the time derivative of energy...

force is the "distance" derivative of energy...

Energy is force*distance... not force*time...

suppose 2 objects have the same mass and acceleration.

one goes from 0m/s to 5m/s. the other goes from 5m/s to 10m/s.

both experience the same net force... and this change in velocity occurs in the same time... however the second object travels a further distance... hence it has more work done on it.

 

[ZapperZ]

That is why I gave an example where 'it appears to me' that conservation of energy is violated ...

But I don't see it.

All you can say is that the amount of energy being used up by the fuel goes to the increase in KE of the ship. That's it. You are insisting here that this energy goes into the SAME change in velocity, rather than the same change in square of the velocity. There's no rational reason why that needs to be so. If you look carefully, it is v_f^2 - v_i^2 that remains constant, not v_f - v_i, over the same period of time.

Zz.

 

[jtbell]

Why does it take more energy to accelerate from 5 to 10 m/s than from 0 to 5 m/s ?

Assume the same constant force in both cases, therefore the same acceleration. Therefore both cases take the same amount of time. In the first case, the average speed is higher, so the object travels further than in the second case. Therefore, W = f.d is larger in the first case.

 

[pixel01]

I don't doubt this is true, I just don't understand it. Why does it take more energy to accelerate from 5 to 10 m/s than from 0 to 5 m/s ? Are these not the same just in a different frame of reference ?

The kinetic energy has squared v in its calculation as explained above, so as a matter of course, energy to accelerate from 0m/s to 5m/s must be different from 5m/s to 10m/s.

 

[TVP45]

The way i have learned to think about this kind of problem goes back to Clerk Maxwell's dictum to always change all energies to a moving mass before doing any sort of comparison. I find that gets me out of this whole confusion of why does a gallon of gas in one frame not produce the same energy as a gallon of gas in another frame. So, instead of supplying fuel to some engine, I bring in energy in the form of a moving mass from an outside frame (that way, frame recoil momentum goes away) and add it as needed in the form of an elastic collision. Then, from what ever frame I choose, I can measure that added energy, and it will be different for each frame.

 

[Emanresu]

All you can say is that the amount of energy being used up by the fuel goes to the increase in KE of the ship. That's it. You are insisting here that this energy goes into the SAME change in velocity, rather than the same change in square of the velocity.
Zz.

But the reason I say that it goes into the same change in velocity is because it is constant acceleration caused by a constant force caused by a constant supply of energy.

I am burning fuel (releasing energy) at a constant rate in time to create a constant force and yet my rocket is not gaining energy at a constant rate in time. So what is happening in terms of conservation of energy ?

Are we agreed that a constant supply of energy can / will generate constant thrust / force ?

E.

 

[Emanresu]

The way i have learned to think about this kind of problem goes back to Clerk Maxwell's dictum to always change all energies to a moving mass before doing any sort of comparison. I find that gets me out of this whole confusion of why does a gallon of gas in one frame not produce the same energy as a gallon of gas in another frame. So, instead of supplying fuel to some engine, I bring in energy in the form of a moving mass from an outside frame (that way, frame recoil momentum goes away) and add it as needed in the form of an elastic collision. Then, from what ever frame I choose, I can measure that added energy, and it will be different for each frame.

I'm glad at least one person doesn't think I'm mad / stupid, even if I didn't fully understand your explanation.

 

[ZapperZ]

But the reason I say that it goes into the same change in velocity is because it is constant acceleration caused by a constant force caused by a constant supply of energy.

I am burning fuel (releasing energy) at a constant rate in time to create a constant force and yet my rocket is not gaining energy at a constant rate in time. So what is happening in terms of conservation of energy ?

Why do you say this? Where is this "increase" in energy?

Let's say the fuel burns an amount of E per unit time.

Now let's do this 2 ways, via a simple conservation of energy and then via the amount of work done by the force being applied. Already we can see that

E = \Delta(KE)

which means that

E = 1/2 m(v_f^2 - v_i^2)

This E is a constant throughout, and so is the difference in the square of the velocity, NOT the difference in velocity. That's why you were making a mistake when you comparing the change in going from 0-5 m/s versus 10-15 m/s. This is NOT the fundamental quantity that you should be dealing.

Now, let's consider this via the force applied, since you were worried about the acceleration. Rewriting the first equation using work done

E = \int{F ds}

where ds is the differential displacement. Now, it has already been pointed out that we can write F as

F = ma = m dv/dt = m \frac{dv}{ds} \frac{ds}{dt} = m \frac{dv}{ds} v

This means that

E = m\int{v dv}

Guess what? When you do the integral, you end up with the exact same expression as the one I got using conservation of energy.

Again, there's gain or change in energy over the unit time period in which E is burned from the fuel. What you are confusing with here is the insistance that \Delta(v) be constant, where as in actuality, it is \Delta(v^2), which is a direct result from the conservation of energy.

So there are no energy being created nor loss here.

Zz.

 

[TVP45]

I'm glad at least one person doesn't think I'm mad / stupid, even if I didn't fully understand your explanation.

No, I think you're looking hard for a way to understand this. And, you should know that many people, many of them with good technical educations have trouble with this same point. The tricky thing to get hold of is that momentum (kg-m/s) is conserved, and energy (kg-m^2/s^2) is conserved, but velocity (m/s) is not. And, if you think about it a little bit, you might see that, in order to conserve momentum for example, you have to let go of trying to conserve velocity.

But, the main thing is that, in all external frames, the energy of the fuel you burn will be measured differently depending on whether you're going from 0 to 5 m/s as opposed to going from 15 to 20 m/s. And the important word there is measured.

Do you know about Galilean transformations? Here's a good link that has some explanations. Some of the other posters have been trying to get you to see that, if you are already moving when you start the acceleration, you will travel farther during the time you accelerate. You can see that in the hyperphysics link.
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

 

[phredbo]

Ek=1/2mvsqu

use the formula for kinetic energy. Ek=1/2m(v)squ] put in any number for [m] and any for [v]. run the formula and find [Ek] take that num and V/Ek and u will find it is proportional. all the best ..phredbo

 

[Emanresu]

E = \Delta(KE)

which means that

E = 1/2 m(v_f^2 - v_i^2)

This E is a constant throughout, and so is the difference in the square of the velocity, NOT the difference in velocity. That's why you were making a mistake when you comparing the change in going from 0-5 m/s versus 10-15 m/s. This is NOT the fundamental quantity that you should be dealing.

Now, let's consider this via the force applied, since you were worried about the acceleration. Rewriting the first equation using work done

E = \int{F ds}

where ds is the differential displacement. Now, it has already been pointed out that we can write F as

F = ma = m dv/dt = m \frac{dv}{ds} \frac{ds}{dt} = m \frac{dv}{ds} v

This means that

E = m\int{v dv}

Zz.

I am not disputing any of this. My big hangup is to do with energy = force x distance, so starting any answer with this fact (which I know is true even though it confuses me) is not going to help me.

Say my rocket has mass of 1kg and converts energy E per second to produce 1N of thrust.

In the first second E will produce a change in KE of 0.5 (0.5 - 0)
In the second second E will produce a change in KE of 1.5 (2 - 0.5)
In the third second E will produce a change in KE of 2.5 (4.5 - 2)
etc.

I'm using the same amount of E (fuel) every second to produce ever increasing amounts of KE (per second).

I am confused.
You are confused as to why I am confused.
I am confused as to why you are confused as to why I am confused.
etc.

E.

 

[ZapperZ]

I am not disputing any of this. My big hangup is to do with energy = force x distance, so starting any answer with this fact (which I know is true even though it confuses me) is not going to help me.

Say my rocket has mass of 1kg and converts energy E per second to produce 1N of thrust.

In the first second E will produce a change in KE of 0.5 (0.5 - 0)
In the second second E will produce a change in KE of 1.5 (2 - 0.5)
In the third second E will produce a change in KE of 2.5 (4.5 - 2)
etc.

I'm using the same amount of E (fuel) every second to produce ever increasing amounts of KE (per second).

I am confused.
You are confused as to why I am confused.
I am confused as to why you are confused as to why I am confused.
etc.

E.

Eh? Can you please show me the ALGEBRAIC expression first on how you do this? I don't know what these numbers represent. If you're still in school, you should know that this is not a very good idea, i.e. to plug in number way in the beginning of the problem rather than solving it symbolically first till the very end.

And not only that, but I still think that you're "hung up" on insisting that the change in velocity per unit time remains constant. I still want to see where that is a requirement when you start from conservation of energy principles. Note that Newton's laws and Lagrangian/Hamiltonian mechanics ALL start from that principle and the principle of conservation of momentum. So those laws can't contradict what it started with as a valid assumption.

In addition, I have already SHOWED you what you will get using the work done by the thrust. I explicitly showed the 2nd method to solving this using "force" via the work done on the vehicle. So I don't know where you got your "expression" from.

Zz.

 

[TVP45]

I just had a thought how you might see this. Consider raising a mass up in the air and then dropping it. If you were to use a little gasoline powered hoist, it realistically would take maybe 100 ml of fuel for each meter raised, so you can equate fuel to height. Then think about dropping it from different heights (work it out on paper) and see how fast it is going after each meter. (Ignore air friction). You'll always gain the same amount of energy each meter, but the amount of ADDED velocity gets smaller because you're falling through each successive meter faster and faster, that is, you're not there long enough to experience much acceleration (remember acceleration is tied to time)

 

[learningphysics]

I am not disputing any of this. My big hangup is to do with energy = force x distance, so starting any answer with this fact (which I know is true even though it confuses me) is not going to help me.

Say my rocket has mass of 1kg and converts energy E per second to produce 1N of thrust.

The above is impossible... constant rate of fuel consumption cannot produce a constant force.

In the first second E will produce a change in KE of 0.5 (0.5 - 0)
In the second second E will produce a change in KE of 1.5 (2 - 0.5)
In the third second E will produce a change in KE of 2.5 (4.5 - 2)
etc.

The above is assuming that a constant rate of fuel consumption produces a constant force (and hence constant acceleration). This isn't right. You can calculate the force on the rocket in this situation if you want... it is a changing decreasing force...

Constant fuel consumption does not imply a constant force.

 

[ZapperZ]

The above is impossible... constant rate of fuel consumption cannot produce a constant force.



The above is assuming that a constant rate of fuel consumption produces a constant force (and hence constant acceleration). This isn't right. You can calculate the force on the rocket in this situation if you want... it is a changing decreasing force...

Constant fuel consumption does not imply a constant force.

That is an excellent point and something the OP has missed a few times. I think when I wrote that

E = \int{F ds}

the significance of that expression is somehow lost, which is exactly what you have said here that the energy generated by the fuel does not necessarily produce a constant force.

Zz.

 

[Ahmed Abdullah]

I am not disputing any of this. My big hangup is to do with energy = force x distance, so starting any answer with this fact (which I know is true even though it confuses me) is not going to help me.

E.

why mechanical work is defined as W=Force x distance?

Is it axiomatic? Is it so defined because the quantity Fx is always practically found to be conserved? Or some more fundamental reasons behind this?

 

[Emanresu]

The above is impossible... constant rate of fuel consumption cannot produce a constant force.

The above is assuming that a constant rate of fuel consumption produces a constant force (and hence constant acceleration). This isn't right. You can calculate the force on the rocket in this situation if you want... it is a changing decreasing force...

Constant fuel consumption does not imply a constant force.

Thank you for your answer. I accept what you say as a fact. So my confusion is to do with why a constant rate of fuel consumption does not produce a constant force. Can you explain this please ?

E.

 

[jtbell]

Did post #19 in this thread not help at all?

 

[Ahmed Abdullah]

Because Kinetic energy = 1/2mv^2
mechanical energy = Fx
and rate of energy supply P=Fv

The concept of energy is nothing concrete, it is mathematical and the most abstract idea...
you can use the above mentioned formulas and conclude that a constant rate of fuel consumption does not necessarily produce a constant force.

 

[Ahmed Abdullah]

Constant supply of energy means that over a given period of time the quantity 1/2m(vf^2-vi^2) should remain constant but P=Fv suggests that as v increases with the elapsing of time F should decrease.

 

[Ahmed Abdullah]

ANOTHER THING TO NOTE:
Same force F working for the same time doesn't do the same work.

Work =Fs
But S=ut + 1/2 at^2
So work = F (ut + 1/2 at^2)

For different values of u (initial velocity) different amount of work is done by the same force F during the same period of time..

 

[learningphysics]

Thank you for your answer. I accept what you say as a fact. So my confusion is to do with why a constant rate of fuel consumption does not produce a constant force. Can you explain this please ?

E.

Suppose an object is gaining kinetic energy from 0 joules... at a rate of P joules per second where P is a constant... ie:

Energy = Pt
(1/2)mv^2 = Pt

now take the derivative of both sides:

mv*dv/dt = P

mv*a = P

ma = P/v

ie: Force = P/v. as the velocity of the object increases... the force driving it foward gets smaller and smaller...



But what did I use above... that kinetic energy = (1/2)mv^2. But why is kinetic energy this way... as (1/2)mv^2?

I think we need to demonstrate why (1/2)mv^2 is useful... and why energy, kinetic energy and work ideas are useful...

The answer to all this comes from the fact that all the fundamental forces are "conservative"... ie: \int Fdx depends only on the final position and initial position... and not the path in between... or anything else for that matter...

Let's take the gravitational force as an example... for simplicity I'm going to say that the force is mg... take an object moving under the influence of a gravitaitonal force and nothing else... so we haven't defined energy yet... we're just working with forces and kinematics...

taking up as positive... down as negative... as an object goes from h1 to h2... \int Fdx is -mg(h2-h1).

see that the result of this integral depends only on h2 and h1... not on the time it takes the object to get from h1 to h2... not on the path between h1 and h2...

now we also know that (1/2)mvf^2 - (1/2)mvi^2 = \int Fdx

so: (1/2)mvf^2 - (1/2)mvi^2 = -mg(h2-h1).

so this is a critical point... the change in velocity is the result of a change in position... or rather, given the initial position and initial velocity... we can calculate the velocity of the object at any time, knowing its position at that time. notice... time is not anywhere here...

we also see that

(1/2)mvf^2 +mg(h2) = (1/2)mvi^2 + mg(h1)

so (1/2)mv^2 + mgh is a conserved quantity! we can find the velocity at any height that this object is at given the initial (1/2)mv^2 + mgh value.

So now, it makes sense to define these terms to make use of that the fact that this quantity is conserved... Kinetic energy as (1/2)mv^2... gravitational potential energy as mgh... total energy as (1/2)mv^2 + mgh.

so:

1) \int Fdx is a useful integral in nature... because the fundamental forces are conservative and this result depends on only initial and final positions... and it gives us a relationship between the initial and final velocities of an object under the influence of this force.

2) Since \int Fdx = (1/2)mvf^2 - (1/2)mvi^2... and since \int Fdx = U(final position) - U(initial position) for some function U... we get conservation of (1/2)mv^2 + U.

3) For the above 2 reasons, it is useful to define a quantity called kinetic energy as (1/2)mv^2... and a quantity U as gravitational potential energy (for our case above)... and the conserved quantity "energy" as the sum of these 2 (or in general sum of kinetic energy and all potential energies)... Also, as a result of defining kinetic energy in this way \int Fdx is hence a change in kinetic energy.

Now, above I was just talking about 1 object just experiencing a gravitational force... but the thing is all the fundamental forces are conservative... the integral of Fdx for any of them depends only on final position and initial position. we can get potential energies for each fundamental force.

Hope this gives some idea of why we define KE the way we do... and why energy is defined the way it is...

 

[Emanresu]

Hope this gives some idea of why we define KE the way we do... and why energy is defined the way it is...

Thanks very much, I'm starting to get my head round it now.

 

[D H]

There are so many things wrong here I almost don't know where to start.

A propulsion system that provides a constant force would typically use energy at a constant rate in order to accelerate at a constant rate ...

I'll start here at the original post. A constant force does not result in a constant acceleration because the vehicle is ejecting mass. Instead, a constant force results in an ever increasing acceleration as the vehicle becomes less massive.

The above is impossible... constant rate of fuel consumption cannot produce a constant force.

Constant rate of fuel production does produce a constant force.

If you burn fuel at a constant rate (kg/sec or liters/sec or whatever) and thereby produce energy at a constant rate (joules/sec), I guarantee that your rate of acceleration will not remain constant, but will decrease as time passes, so that the kinetic energy increases at a constant rate. Specifically, v as a function of time will be proportional to the square root of t.

If you burn fuel at a constant rate, I guarantee that your rate of acceleration will increase as time passes. As an example, during second stage ascent, the Shuttle has to throttle back the main engines to keep acceleration under 3g as the vehicle loses mass. The astronauts would not be able to handle the increasing acceleration if the main engines were fired at maximum thrust.

-----------------

The main problem is that everyone has ignored the exhaust.

Let's put the rocket in deep space, far from any masses. The rocket and all of the exhaust form a closed system: Momentum is conserved. The momentum of the rocket is

p_r = m_r(t)v_r(t)

Differentiating wrt time,

\frac{d p_r(t)}{dt} = \frac{d m_r(t)}{dt}v_r(t) + m_r(t)a_r(t)

The rocket adds momentum to the exhaust cloud as

\frac{d p_e(t)}{dt} = \frac{d m_e(t)}{dt} (v_r(t)+v_e(t))

where v_e(t)[/tex] is the exhaust velocity relative to the vehicle.<br /> By conservation of mass and conservation of momentum,<br /> <br /> \frac{d m_e(t)}{dt} = -\,\frac{d m_r(t)}{dt}<br /> \frac{d p_e(t)}{dt} = -\,\frac{d p_r(t)}{dt}<br /> <br /> Simplifying,<br /> <br /> \frac{d m_r(t)}{dt} v_e(t) = m_r(t)a_r(t)<br /> <br /> Let the rate at which fuel is consumed be denoted as \dot m_f(t)[/tex]. Then \dot m_e(t) = -\dot m_r(t) = \dot m_f(t)[/tex] and&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; a_r(t) = -\,\frac{\dot m_f(t)}{m_r(t)} v_e(t)&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; What then about energy? The rocket at any point in time has kinetic energy&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; E_r(t) = \frac 1 2 m_r(t)v_r(t)^2&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; Assuming a constant relative exhaust velocity, the change in the rocket&amp;amp;#039;s energy as it acquires some change in velocity \Delta v (and loses mass) is&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; \Delta E_r(t) = \frac 1 2 m_f\left(e^{\Delta v/v_e}-1)\v_e^2&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; A little more work yields the Tsiolkovsky rocket equation,&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; \Delta v = v_e\ln\frac{m_{\text{init}}}{m_{\text{final}}}&amp;lt;br /&amp;gt; The change in energy is also proportional to the quantity of fuel burned.

 

[D H]

Just a bit more math to show the change in total kinetic energy is independent of the rocket velocity. I'll keep this to one dimension.

Let
v_r be the velocity of the rocket
v_e[/itex] be the exhaust velocity relative to the rocket<br /> \dot m[/itex] be the fuel consumption rate&lt;br /&gt; \Delta t[/itex] be some small time interval&amp;lt;br /&amp;gt; m_r[/itex] be the mass of the rocket and fuel on the rocket at the start of the interval&amp;amp;lt;br /&amp;amp;gt; \Delta m[/itex] be the mass of the exhaust ejected over the interval.&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; At the start of the time interval, the rocket has momentum&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; p_{r,0} = m_r v_r&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; The momentum of the rocket at the end of the interval is&amp;amp;amp;lt;br /&amp;amp;amp;gt; p_{r,f} = (m_r-\Delta m) (v_r+\Delta v_r)&amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; \approx m_r v_r + m_r \Delta v_r - \Delta m v_r&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; The change in the rocket exhaust is thus&amp;amp;amp;lt;br /&amp;amp;amp;gt; \Delta p_r =m_r \Delta v_r - \Delta m v_r&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; The exhaust has gained momentum&amp;amp;amp;lt;br /&amp;amp;amp;gt; \Delta p_e = \delta m (v_r-v_e) = \delta m v_r - \delta m v_e&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; The total change in momentum for a closed system (rocket+exhaust) is zero:&amp;amp;amp;lt;br /&amp;amp;amp;gt; \Delta p_r + \Delta p_e = m_r \Delta v_r - \Delta m v_e = 0&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; or&amp;amp;amp;lt;br /&amp;amp;amp;gt; \Delta v_r = \frac{\Delta m}{m_r} v_eAt the start of the time interval, the rocket has kinetic energy&amp;amp;amp;lt;br /&amp;amp;amp;gt; p_{r,0} = \frac 1 2 m_r v_r^2&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; The kinetic energy of the rocket at the end of the interval is&amp;amp;amp;lt;br /&amp;amp;amp;gt; p_{r,f} = \frac 1 2 (m_r-\Delta m) (v_r+\Delta v_r)^2&amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; \approx \frac 1 2 m_r v_r^2 - \frac 1 2 \Delta m_r v_r^2 + m_r v_r \Delta v_r&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; The change in the rocket exhaust is thus&amp;amp;amp;lt;br /&amp;amp;amp;gt; \Delta E_r = m_r v_r \Delta v_r - \frac 1 2 \Delta m_r v_r^2&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; Applying the result from conservation of momentum, m_r \Delta v_r = \Delta m v_e,&amp;amp;amp;lt;br /&amp;amp;amp;gt; \Delta E_r= \Delta m v_r v_e - \frac 1 2 \Delta m_r v_r^2&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; The exhaust has gained kinetic energy&amp;amp;amp;lt;br /&amp;amp;amp;gt; \Delta E_e = \frac 1 2 \Delta m (v_r-v_e)^2&amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; = \frac 1 2 \Delta m v_r^2 - \Delta m v_r v_e + \frac 1 2 \Delta m v_e^2&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; The total change in kinetic energy is&amp;amp;amp;lt;br /&amp;amp;amp;gt; \Delta E_r + \Delta E_e = \frac 1 2 \Delta m v_e^2&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; which is independent of the rocket velocity.

 

[ZapperZ]

But as I've mentioned to you via PM, and by the fact that "m" has been taken to be a constant here, the rate of change of mass of the vehicle itself is not being considered here. One could easily use an external source to impart the "energy" to the vehicle. Recall the original premise here:

Could someone please explain why kinetic energy is proportional to velocity squared and not just velocity.

A propulsion system that provides a constant force would typically use energy at a constant rate in order to accelerate at a constant rate and thus rate of change of velocity would be proportional to rate of use of energy. So why is rate of gain of energy not also proportional to rate of change of velocity ?

If you notice the very first line, the OP is confused about why KE is proportional to v^2 and not v, which is why I was pointing out that the important issue here is the difference in the square of the velocity, not the difference in the velocity. So this isn't the regular "rocket" problem that not only has a dv/dt term, but also the dm/dt term associated with the force.

At least, that is how I saw it as the central issue. The rocket is simply being used to illustrate the central issue. Of course, the original poster can always explain if the change in mass is part of the issue. That part, I haven't seen.

Zz.

 

[russ_watters]

It is easier to see with a car than a rocket since people deal with them every day (and their mass doesn't change as they accelerate). Cars accelerate from 0-30 much, much faster than from 30-60. And even with roughly constant horsepower, you can see that the reason is that the force (the torque at the wheels) decreases as you go through theh gears.

 

[russ_watters]

The concept of energy is nothing concrete, it is mathematical and the most abstract idea...

You have it backwards. The math is what makes the concept absolutely concrete. Math is precise. I don't understand why people tend to want to believe there is a deeper meaning than that (especially an abstract one).

 

[D H]

If you simply eliminate the constant acceleration stuff the OP was right. A propulsion system that produces a constant force does indeed burns fuel (uses energy) at a constant rate. Yes, the OP was also confused with the v² term in a kinetic energy. This is one of the things I strongly object to here:

That is why I gave an example where 'it appears to me' that conservation of energy is violated ...

But I don't see it.

All you can say is that the amount of energy being used up by the fuel goes to the increase in KE of the ship. That's it. You are insisting here that this energy goes into the SAME change in velocity, rather than the same change in square of the velocity. There's no rational reason why that needs to be so. If you look carefully, it is v_f^2 - v_i^2 that remains constant, not v_f - v_i, over the same period of time.

Zz.

The reason that conservation of energy appears to fail is because all of you are forgetting about the exhaust. Forgetting about the exhaust leads to statements like this (names omitted)

The above is impossible... constant rate of fuel consumption cannot produce a constant force.

and this

If you burn fuel at a constant rate (kg/sec or liters/sec or whatever) and thereby produce energy at a constant rate (joules/sec), I guarantee that your rate of acceleration will not remain constant, but will decrease as time passes, so that the kinetic energy increases at a constant rate. Specifically, v as a function of time will be proportional to the square root of t.

both of which are so flipping WRONG that I cannot restrain from shouting.

 

[ZapperZ]

If you simply eliminate the constant acceleration stuff the OP was right. A propulsion system that produces a constant force does indeed burns fuel (uses energy) at a constant rate. Yes, the OP was also confused with the v² term in a kinetic energy. This is one of the things I strongly object to here:



The reason that conservation of energy appears to fail is because all of you are forgetting about the exhaust. Forgetting about the exhaust leads to statements like this (names omitted)



and this



both of which are so flipping WRONG that I cannot restrain from shouting.

Well maybe you need to calm down a bit and figure out where most of are approaching this problem and see if the OP actually is simply using the rocket to illustrate an issue, or if this REALLY is a rocket problem. We could easily use an external source to impart that same energy to the rocket (i.e. I could have a constant flux of particle bombarding the vehicle from behind that doesn't stick to it).

I would say that most of us here are VERY familiar with the "rocket" problem in which dm/dt has to be considered. So if this is true, and if you think that most of us here are not morons, why do you think we never included that in THIS problem?

Zz.

 

[D H]

We could easily use an external source to impart that same energy to the rocket (i.e. I could have a constant flux of particle bombarding the vehicle from behind that doesn't stick to it).

Fine. Outfit our vehicle with a light sail. We'll power the light sail from a "fixed" source. Keeping this to one dimension once again for simplicity, and keeping velocities small to avoid relativistic effects, the momentum transferred to the vehicle over some time interval is \Delta p = 2 \dot n h/\lambda \Delta t, where \dot n is the photon flux. The factor of two results from the reflecting the incoming photon beam back toward the transmitter. The energy transferred to the vehicle is \Delta E = 2 \dot nh c/\lambda \Delta t, obviously proportional to the change in momentum. Since the mass is constant, the change in vehicle velocity is also proportional to energy transferred to the vehicle.

Suppose this were not the case: the change in velocity depends on the initial velocity. This means an observer in a inertial frame initially at rest with respect to the vehicle will see a different change in velocity than an observer in an inertial frame in which the vehicle has some non-zero initial velocity. As the relative velocity of the two observer frames is constant, both observers should see the same change in velocity.

 

[ZapperZ]

OK, bad example that I came up with on the fly. That obviously is a constant force source.

Still, I don't see where you can find the flaw if the original premise if don't include any dm/dt term. Shouldn't we wait for the OP to come back and clarify if this is what he/she want to include? I mean, after all, THIS is where we differ in interpretation of the original intent of the question.

Zz.

 

[animalcroc]

I think the problem is that everyone is using calculus to explain KE. I'll use simlple algebra.
First though, you should note that KE is defined to be .5mv^2 and momentum is defined to be mv. These are distinct but equally important concepts.

Now, to define or derive KE using constant acceleration:
KE=fx=max=m [(Vf-Vi)/t] [(Vf+Vi)/2]=(1/2)[Vf^2-Vi^2]=(1/2)Vf^2 taking Vi=0

Just as KE=fx is conserved, another quantity Impulse=ft is also conserved. You may know that impulse=change in momentum. See how similary impulse and KE are? Distance versus time.

 

[ZapperZ]

I think the problem is that everyone is using calculus to explain KE. I'll use simlple algebra.
First though, you should note that KE is defined to be .5mv^2 and momentum is defined to be mv. These are distinct but equally important concepts.

Er.. you need to re-read what has been written here. The reason why you can do that is the very reason why you can't assume that acceleration is a constant. There's nothing here that tells you that because you don't know that F is a constant to start with. The ONLY constant here is the energy transferred to the vehicle. As you can already see from the thread, this does not automatically imply a constant force. So you can't simply assume that acceleration is a constant.

You should also consider that using "calculus" is more general that what you have done here, which is more of a specific case. the use of calculus isn't the issue nor the cause of any of the disagreement here.

I suggest we WAIT for the OP to get back and clarify the exact scenario of the problem at hand.

Zz.