Why kinetic energy is proportional to velocity squared

[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 [itex]v_e(t)[/tex] is the exhaust velocity relative to the vehicle.
By conservation of mass and conservation of momentum,

$$\frac{d m_e(t)}{dt} = -\,\frac{d m_r(t)}{dt}$$
$$\frac{d p_e(t)}{dt} = -\,\frac{d p_r(t)}{dt}$$

Simplifying,

$$\frac{d m_r(t)}{dt} v_e(t) = m_r(t)a_r(t)$$

Let the rate at which fuel is consumed be denoted as [itex]\dot m_f(t)[/tex]. Then [itex]\dot m_e(t) = -\dot m_r(t) = \dot m_f(t)[/tex] and

$$a_r(t) = -\,\frac{\dot m_f(t)}{m_r(t)} v_e(t)$$

What then about energy? The rocket at any point in time has kinetic energy

$$E_r(t) = \frac 1 2 m_r(t)v_r(t)^2$$

Assuming a constant relative exhaust velocity, the change in the rocket's energy as it acquires some change in velocity $\Delta v$ (and loses mass) is

$$\Delta E_r(t) = \frac 1 2 m_f\left(e^{\Delta v/v_e}-1)\v_e^2$$

A little more work yields the Tsiolkovsky rocket equation,

$$\Delta v = v_e\ln\frac{m_{\text{init}}}{m_{\text{final}}}$$
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
[tex]v_e[/itex] be the exhaust velocity relative to the rocket
[tex]\dot m[/itex] be the fuel consumption rate
[tex]\Delta t[/itex] be some small time interval
[tex]m_r[/itex] be the mass of the rocket and fuel on the rocket at the start of the interval
[tex]\Delta m[/itex] be the mass of the exhaust ejected over the interval.

At the start of the time interval, the rocket has momentum

$$p_{r,0} = m_r v_r$$

The momentum of the rocket at the end of the interval is
$$p_{r,f} = (m_r-\Delta m) (v_r+\Delta v_r) \approx m_r v_r + m_r \Delta v_r - \Delta m v_r$$

The change in the rocket exhaust is thus
$$\Delta p_r =m_r \Delta v_r - \Delta m v_r$$

The exhaust has gained momentum
$$\Delta p_e = \delta m (v_r-v_e) = \delta m v_r - \delta m v_e$$

The total change in momentum for a closed system (rocket+exhaust) is zero:
$$\Delta p_r + \Delta p_e = m_r \Delta v_r - \Delta m v_e = 0$$

or
$$\Delta v_r = \frac{\Delta m}{m_r} v_e$$At the start of the time interval, the rocket has kinetic energy
$$p_{r,0} = \frac 1 2 m_r v_r^2$$

The kinetic energy of the rocket at the end of the interval is
$$p_{r,f} = \frac 1 2 (m_r-\Delta m) (v_r+\Delta v_r)^2 \approx \frac 1 2 m_r v_r^2 - \frac 1 2 \Delta m_r v_r^2 + m_r v_r \Delta v_r$$

The change in the rocket exhaust is thus
$$\Delta E_r = m_r v_r \Delta v_r - \frac 1 2 \Delta m_r v_r^2$$

Applying the result from conservation of momentum, $m_r \Delta v_r = \Delta m v_e$,
$$\Delta E_r= \Delta m v_r v_e - \frac 1 2 \Delta m_r v_r^2$$

The exhaust has gained kinetic energy
$$\Delta E_e = \frac 1 2 \Delta m (v_r-v_e)^2 = \frac 1 2 \Delta m v_r^2 - \Delta m v_r v_e + \frac 1 2 \Delta m v_e^2$$

The total change in kinetic energy is
$$\Delta E_r + \Delta E_e = \frac 1 2 \Delta m v_e^2$$

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.

 

[learningphysics]

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



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.



Constant rate of fuel production does produce a constant force.

I apologize. Yes, what I wrote was wrong. I was thinking in very simple terms of an object with a fixed mass gaining kinetic energy at a constant rate...

Sorry Emanresu for any confusion I caused.

 

[learningphysics]

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.

Yeah, car would have worked, since it isn't driven by exhaust, but by the force of the tires on the road...

 

[animalcroc]

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.

Of course calculus is more general but to explain a concept you start with the basics so it's more understandable. In fact, we don't even know the math level of the OP.
Intro physics book also use the constant acceleration approach. Many concepts are introduced without calculus because the math levels of readers are unknown and a better intuitive understanding may be developed without using calculus.
I only answered the first question of the OP (which was clear).

 

[Emanresu]

I apologize. Yes, what I wrote was wrong. I was thinking in very simple terms of an object with a fixed mass gaining kinetic energy at a constant rate...

Sorry Emanresu for any confusion I caused.

No problem. What you explained earlier made sense. But I am confused again as I didn't follow DH's argument. If now 'Constant rate of fuel production does produce a constant force' I am back to square one. Can you explain DH's argument for me ?

E.

 

[Emanresu]

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.

You are a bit over my head here, but when you say 'Since the mass is constant, the change in vehicle velocity is also proportional to energy transferred to the vehicle', is there a constant force on the vehicle ?

E.

 

[D H]

Emanresu, we are having a bit of an issue with regard to your intent in your original post. As you asking about why energy is proportional to the square of velocity rather than velocity and using rockets to illustrate the issue, or are you asking about rockets and wondering about energy as a tangential issue, or both?

 

[Emanresu]

Emanresu, we are having a bit of an issue with regard to your intent in your original post. As you asking about why energy is proportional to the square of velocity rather than velocity and using rockets to illustrate the issue, or are you asking about rockets and wondering about energy as a tangential issue, or both?

Originally I thought that constant rate of energy supply would provide constant force so I was confused as to why energy out (KE) was proportional to v^2. Then as result of answers given here I was persuaded that a constant force would not result which explained why KE was not proportional to v. But you are saying something different which is similar to what I originally thought (constant energy supply WILL provide constant force) which in my mind does not tally with KE being proportional to v^2. I'm obviously not very good at explaining things, hence all the confusion, but if you could provide a simplified explanation of your light sail example, maybe that would help me.

E.

p.s. I only used a rocket because I thought it was the simplest concept (I was obvioulsy wrong).

 

[D H]

A rocket is far from the simplest concept. Let me pick at another post of yours:

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 ?

You are missing the exhaust. Several things are going on here. First, a thrusting rocket loses mass. If the rocket does manage to generate constant thrust its acceleration will increases.

So, what about energy?

Examine what happens to the rocket and exhaust cloud over some small time interval $\Delta t$. (By small I mean sufficiently short in duration such that products of changes can be ignored). I am assuming the rocket is well removed from any massive bodies so I can ignore gravitational potential energy.

At the start of the interval, the rocket has kinetic energy $E_{r-} = \frac 1 2 m_r v_r^2$. Denote the initial kinetic energy of the exhaust cloud as some unknown and irrelevant quantity $E_{e-}$. Over the time interval, the rocket ejects $\Delta m$ quantity of fuel as exhaust moving at a velocity $v_e$ relative to the vehicle. The kinetic energies of the rocket and exhaust cloud at the end of the interval are

$$E_{r+} = \frac 1 2 (m_r-\Delta m)(v_r+\Delta v_r)^2 = E_{r-} + m_r v_r\cdot\Delta v_r - \frac 1 2 \Delta m v_r^2 + O(\Delta^2)$$
$$E_{e+} = E_{e-} + \frac 1 2 \Delta m(v_r+v_e)^2 = E_{e-} + \frac 1 2 \Delta m v_r^2 + \Delta m v_r\cdot v_e + \frac 1 2 \Delta m v_e^2$$

The total kinetic energy is thus
$$E_{+} = E_{-} + m_r v_r\cdot\Delta v_r - \frac 1 2 \Delta m v_r^2 + \frac 1 2 \Delta m v_r^2 + \Delta m v_r\cdot v_e + \frac 1 2 \Delta m v_e^2+ O(\Delta^2)$$

The change in kinetic energy is just $E_{+} - E_{-}$. Eliminating high order terms and simplifying,
$$\Delta E = m_r v_r\cdot\Delta v_r + \Delta m v_r\cdot v_e + \frac 1 2 \Delta m v_e^2$$

The rocket and exhaust cloud form a closed system subject to conservation of linear momentum. Applying conservation of momentum yields $m_r \Delta v_r + \Delta m v_e=0$. Applying this result to the change in kinetic energy yields
$$\Delta E = \frac 1 2 \Delta m v_e^2$$

Note that this result is independent of the rocket velocity. This quantity represents the useful energy obtained by burning fuel and ejecting exhaust from the vehicle. This quantity will alwys be less than the chemical energy in the fuel, as some of the chemical energy is lost in the form of an increased exhaust temperature.

 

[learningphysics]

No problem. What you explained earlier made sense. But I am confused again as I didn't follow DH's argument. If now 'Constant rate of fuel production does produce a constant force' I am back to square one. Can you explain DH's argument for me ?

E.

A rocket... or an "exhaust-based' propulsion is much more complicated... since the mass of the fuel/exhaust ejected is significant (it has to be since the method of propulsion is shooting out exhaust... the greater the momentum of the exhaust that is shot out... ie the more exhaust is shot out... the greater the change in momentum of what's left of the rocket... due to conservation of momentum) And

mass of the exhaust can't be ignored... so the mass of the rocket is changing with time... and a significant portion of the energy in the fuel is lost with the exhaust.

As DH wrote, a constant exhaust velocity along with a constant rate of fuel consumption, leads a constant force on the rocket (whose mass is dropping)...

I don't think the rocket example is the best example to use to get across the idea that kinetic energy is proportional to v^2. A car (normal everyday car, not a rocket-type one) is a better example as russ mentioned... a car isn't driven forward by the exhaust but by the force of the tires against the road... the mass of the fuel and exhaust is very small compared to the mass of the car so the change in mass can be ignored...


The main idea is that an object with a fixed mass, gaining kinetic energy at a constant rate does not experience a constant force (rocket is not an example of such a situation since mass is changing)... what I posted before applies to this idea of a fixed mass and constant rate of kinetic energy (P) increase... I'll post it here too:

after a time t, the object has kinetic energy:

E = (1/2)mv^2
Pt = (1/2)mv^2
$$v = \sqrt{\frac{2Pt}{m}}$$

taking the derivative of Pt = (1/2)mv^2:

P = mv*dv/dt

P = mv*a

P = (ma)*v

P = F*v

F = P/v.

$$F = \frac{P}{\sqrt{\frac{2Pt}{m}}}$$

So F the force on the object changes with time...

I think the thread may have gone a little aside from your questions... I think that your main questions are rooted in why energy is defined as it is... why (1/2)mv^2, why $$\int Fdx$$ etc... Hope my post #38 gives some idea of these ideas... the essential idea that leads to these I believe is that the fundamental forces are conservative...

 

[Emanresu]

I think the thread may have gone a little aside from your questions... I think that your main questions are rooted in why energy is defined as it is... why (1/2)mv^2, why $$\int Fdx$$ etc... Hope my post #38 gives some idea of these ideas... the essential idea that leads to these I believe is that the fundamental forces are conservative...

Thanks again. I'm happy with the principle now, although I'm going to have a go at working through DH's rocket math just out of interest.

E.

 

[D H]

No one peer reviewed what I wrote. No guarantees on sign errors, etc. I wrote it from scratch. (BTW, modeling spacecraft dynamics is one of my jobs.)

Google "Tsiolkovsky rocket equation" for more info. The first several hits are fairly good, including the wikipedia article. Don't bother with the wolfram.com article; it's void of content.

 

[Emanresu]

Note that this result is independent of the rocket velocity. This quantity represents the useful energy obtained by burning fuel and ejecting exhaust from the vehicle. This quantity will alwys be less than the chemical energy in the fuel, as some of the chemical energy is lost in the form of an increased exhaust temperature.

Okay. I've worked through your math and am happy enough with it.

In order to get some insight I started plugging in numbers.

It appears that the faster the rocket is going the more energy is transferred to the exhaust gas

$$\frac 1 2 \Delta m v_r^2 + \Delta m v_r\cdot v_e + \frac 1 2 \Delta m v_e^2$$

because $$v_e$$ is constant and $$v_r$$ is getting bigger.

But that means less energy is being transferred to the rocket as it goes faster which means acceleration is decreasing.

Which means I've made a mistake somewhere (surprise !).

E.

 

[richard14]

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.

in simple words emanresu the velocity is directly proportional to the kinetic energy the biger
the velocity the more kinetic energy obtain by the object let say a car.
was this helpful that's why V2

 

[ZapperZ]

in simple words emanresu the velocity is directly proportional to the kinetic energy the biger
the velocity the more kinetic energy obtain by the object let say a car.
was this helpful that's why V2

This is wrong. KE is proportional the v^2, not v.

Zz.

 

[PatPwnt]

Doesn't 2ad = v^2?

So 2a*d*m = m*v^2 -- a*m = N

2N*d = m*v^2

so N*d = 1/2m*v^2 -- N*d = one of definition of work/energy. Force through a
displacement.

Energy = Energy

a = acceleration d = displacement m = mass

It takes N Newtons through a distance d to stop a mass with velocity v. This is where 1/2mv^2 comes from.

Or... it takes N Newtons through a distance d to accelerate a mass to velocity v.

Can it be any simpler?

 

[ZapperZ]

Doesn't 2ad = v^2?

You should check how that expression was derived. You'll discover that you had implicitly assumed a constant acceleration situation, which may not be true in this situation.

Zz.

 

[rcgldr]

In the non-rocket situation, the frame of reference needs to be non-accelerating reference relative to the point of application of the force. In the case of a car, the point of reference would be the road, or a "fixed" point if you want to include the fact that the Earth is rotated backwards a bit by a car accelerating forwards.

In the case of the rocket, the point of application of the force is the rocket's engine, which is accelerating along with the rocket, accelerating a small part of the rockets mass in one direction (spent fuel) and accelerating the rocket and remaining fuel in the other direction. Note that kinetic energy of both the spent fuel and the rocket (and it's remaining fuel) are significant and part of the "system". As posted below, with a constant burn rate, and constant thrust, the sum of kinetic energy of spent fuel and rocket (and it's remaining fuel) increases linearly with time.

If the spent fuel is ignored, then this approach could be used: As the rocket accelerates, so does the remaining fuel, which increases it's "potential energy". So although the chemical energy remains constant, the total energy per unit of fuel increases with the square of velocity (relative to the starting velocity of the rocket). The net result is that the fuel will supply a fixed thrust per rate of fuel burned, regardless of the velocity of the rocket. The acceleration of spent fuel is always relative to a rocket's current velocity.

 

[D H]

It appears that the faster the rocket is going the more energy is transferred to the exhaust gas

$$\frac 1 2 \Delta m v_r^2 + \Delta m v_r\cdot v_e + \frac 1 2 \Delta m v_e^2$$

because $$v_e$$ is constant and $$v_r$$ is getting bigger.

But that means less energy is being transferred to the rocket as it goes faster which means acceleration is decreasing.

Which means I've made a mistake somewhere (surprise !).

The correct expression is
$$\Delta E = m_r v_r\cdot\Delta v_r + \Delta m v_r\cdot v_e + \frac 1 2 \Delta m v_e^2$$

By conservation of momentum, $m_r \Delta v_r + \Delta m \cdot v_e = 0$, so the change in energy is just

$$\Delta E = \frac 1 2 \Delta m v_e^2$$

=======================

As the rocket accelerates, so does the remaining fuel, which increases it's "potential energy". So although the chemical energy remains constant, the total energy per unit of fuel increases with the square of velocity (relative to the starting velocity of the rocket). The net result is that the fuel will supply a fixed thrust per rate of fuel burned, regardless of the velocity of the rocket.

As the fuel is used up, the mass of the rocket and remaining fuel decrease, and the rate of acceleration increases.

Thank you, Jeff.

The exact same logic also applies to automobiles, which, like rockets, carry the fuel that propels the vehicle. The reason it is harder to accelerate from 30 to 60 MPH than from 0 to 30 in consumer cars is not so much a work issue as it is an increase in aerodynamic drag and rolling friction and because of the relatively low power output of consumer-class automobiles. Racing cars are a different kind of beast. Their engines produce peak torque (and hence peak acceleration) at a much higher engine speed than do consumer-class vehicles.

 

[rcgldr]

To simplify the case of the rocket, imagine that the rocket is held in place (or that the rocket is the frame of reference). The only work done is to the spent fuel, which has a fixed terminal velocity (at a specific burn rate). Kinetic energy of the spent fuel increases linearly with time (velocity of spent fuel is constant, with only the amount of mass of fuel changing linearly with time). If the rocket is not held in place, and a non-accelerating frame of reference is chosen, then the increase in the sum of kinetic energy of spent fuel, and rocket (and it's remaining fuel) will also increase linearly with time, since the source of the work being done, burning of fuel, remains the same. I haven't checked, but the equations shown previously should show this to be true.

In the case of a car, the point of application of force is the pavement and driven tires of the car. This is a simpler case, power = force times speed, for example horsepower = force (lbs) times speed (mph) divided by 375 (conversion factor). Assuming no slippage of the tires, and that the power of the engine is constant, and that a continously variable transmission is used to keep the engine running at maximum power, then force will decrease linearly as speed increases, and kinetic energy will increase at a constant rate. The exhaust from the car is not a significant source of acceleration, and I'm ignoring it in this case.

Note that in both the case of the rocket and the car, kinetic energy of the systems were increasing at a constant rate. The rocket case isn't so special if you take into account the kinetic energy of the spent fuel.

 

[Emanresu]

The correct expression is
$$\Delta E = m_r v_r\cdot\Delta v_r + \Delta m v_r\cdot v_e + \frac 1 2 \Delta m v_e^2$$

By conservation of momentum, $m_r \Delta v_r + \Delta m \cdot v_e = 0$, so the change in energy is just

$$\Delta E = \frac 1 2 \Delta m v_e^2$$

No. I quoted what you gave for the EXHAUST GAS ONLY. Please re-read my question. I know my questions APPEAR stupid to you because you know the subject inside out, but I genuinely am just trying to understand this one bit at a time.

E.