# Physics question

### #1

Posted 04 March 2017 - 23:20

So, kinetic energy has the formula 1/2.m.v^2

So, what about the following scenario. An object weighing a kg is accelerated to 10m/s. In the above formula, this will take 50 joules to achieve.

If the same object is accelerated from rest to 20m/s it would take 200 joules.

Therefore, if the same object is accelerated from 10m/s to 20m/s, shouldn't it take 150 joules to do so? Why does going from to 10 to 20 take so much more energy or am I misunderstanding the formula? Is there an obvious solution?

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### #2

Posted 04 March 2017 - 23:38

Well it's because 20^2-10^2 is much bigger than 10^2-0^2, if you just want the maths.

But, far more interesting is if you look at the definition of work done, Force times the distance that it moves in its own direction. So if we assume a constant force accelerating the object (it doesn't matter, it just makes the maths easy). Let's take F=5 N as an example. Then Newton (2) says F=m.a, so the acceleration is 5 m/s/s, and it will take 2 seconds to accelerate up to 10 m/s, and its average speed is 5 m/s, so it'll cover 10m, so the work done is 5N*10m=50J.

For the next bit again the time taken is 2s. But this time the average speed is 15 m/s, so the distance covered is 2*15=30m. So the work done is 30m*5m=150J

QED

### #3

Posted 05 March 2017 - 02:23

Side note.

This is the reason a vehicle with a given power experiences diminishing acceleration as velocity increases - even in the absence of aerodynamic drag. Power = force x velocity so as velocity increases, force (acceleration) must decrease if power is constant.

### #4

Posted 05 March 2017 - 02:48

Im curious at what do you say is the right answer Ali (other than 150 J, that is). Where is the supposed error?

### #5

Posted 05 March 2017 - 18:17

### #6

Posted 05 March 2017 - 18:26

I find it hard to get my head round to be honest. Speed is a relative rather than an absolute thing. So something that takes 150 joules to go from 10m/s to 20m/s relative to some object is actually going from 0m/s to 10m/s relative to some other object, so from that point of view it should only require 50 joules.

This is what I cannot work out.

Am I understanding the concept of energy incorrectly?

Just say a rocket is in space in a vacuum. It does a burn to go from rest to 10m/s. If it then does a second burn with the exact same quantity of fuel (same energy) does it mean it will not accelerate to 20m/s but to some slower velocity.

I can't get my head around this.

### #7

Posted 05 March 2017 - 20:48

If E means speed v, then 2E leads to 1.41v.

### #8

Posted 05 March 2017 - 21:13

In your example you have changed the reference frames. You started in a reference Frame A and then switched to reference Frame B that is moving at 10 m/s relative to Frame A . Energy conservation laws in newtonian physics are just valid in the same reference frame.

### #9

Posted 05 March 2017 - 22:47

Rockets are different. A rocket throws mass backwards at high velocity (not very efficient because a lot of the power goes into kinetic energy of the exhaust). The thrust and acceleration remain constant until relativistic speeds are approached. So the same burn will achieve the same acceleration regardless of speed.

A car throws the mass of the earth "backwards" which is very efficient but as speed increases, the car has to do work on a mass that is already moving backwards which reduces the force for a given power.

### #10

Posted 05 March 2017 - 22:59

I think a few minutes watching Earl Haig Physics may help you

http://earlhaig.ca/d...ysicsreview.php

particularly the Kinetic Energy and Work videos starting with

https://www.youtube....37546C8&index=1

Yes I know they are High School physics and most of us are way past that but I think you've missed a conceptual step somewhere.

### #11

Posted 06 March 2017 - 00:04

In your example you have changed the reference frames. You started in a reference Frame A and then switched to reference Frame B that is moving at 10 m/s relative to Frame A . Energy conservation laws in newtonian physics are just valid in the same reference frame.

Are you saying that if we ran a car with one side on a conveyor belt going 10 m/s, and the other 2 wheels against the fixed ground, then if the car were decelerated from 20 to 10 m/s using only the brakes on the side rolling against the fixed ground, it would dissipate 3 times more energy than using only the brakes on the conveyor belt side (bringing the car to 0 m/s relative to the conveyor belt)?

**Edited by imaginesix, 06 March 2017 - 00:07.**

### #12

Posted 06 March 2017 - 03:57

Even "simple physics" or "high school physics" is not really simple. It is easy enough to work with the various equations derived from Newton's Laws (he really was a clever bastard) and quote them endlessly as "explanations" - but can you explain to somebody the difference in commonsense terms between momentum and kinetic energy without using any equations?

Newton's First and Second Laws really come down to the property that mass has - inertia. So how to explain the existence of "inertia"? - nobody knows - and really can't even guess at just why mass has inertia.

Also why did Newton decide that F = ma? Why not F=3/4ma or something similar? Surely didn't do any accurate experiments?

### #13

Posted 06 March 2017 - 06:50

F=1000ma (F in Newtons, m in grams, a in m/s/s)

### #14

Posted 06 March 2017 - 09:53

Are you saying that if we ran a car with one side on a conveyor belt going 10 m/s, and the other 2 wheels against the fixed ground, then if the car were decelerated from 20 to 10 m/s using only the brakes on the side rolling against the fixed ground, it would dissipate 3 times more energy than using only the brakes on the conveyor belt side (bringing the car to 0 m/s relative to the conveyor belt)?

Relative to what frame of reference?

That's the point. No matter what side you apply the brakes on, what matters is where you are when you measure it.

Think of you being ran over by that car on either the belt or the floor. Which would throw you higher?

Look here: http://physics.stack...aking-a-vehicle

EDIT: Oh, and about the rocket example, the mass will vary greatly during the process, it can't be simplified and considered constant like that.

It brings another concept, system boundaries, and in this case they are set wrong. You should consider every molecule of water shed by the rocket engine and it's direction and speed on the final energy balance.

**Edited by saudoso, 06 March 2017 - 10:29.**

### #15

Posted 06 March 2017 - 22:22

EDIT: Oh, and about the rocket example, the mass will vary greatly during the process, it can't be simplified and considered constant like that.

The mass change of the rocket seems to be a red herring though, and doesn't directly relate to the problematic scenario. In any case, we can imagine a rocket being accelerated by an outside energy source, such as a laser beam.

If this rocket is accelerated from 0m/s to 10m/s, it takes 50 joules of energy. If it's accelerated from 10m/s to 20/s, it takes 150 joules of energy. But there's no objective fact of the matter in terms of what speed it was going to start with. But there does seem to be an objective fact of how much energy has been used.

I'm not saying that we on this forum have suddenly discovered 0.5mv^2 to be wrong, but I think a lot of people who think that it is simple and that they understand it don't.

Edit - this https://www.physicsf...ncrease.306809/ might be worth looking through.

**Edited by PlatenGlass, 06 March 2017 - 22:42.**

### #16

Posted 06 March 2017 - 22:52

Oh, and about the rocket example, the mass will vary greatly during the process, it can't be simplified and considered constant like that.

Of course it can. You simply consider a small enough increment. Or you could refuel the rocket after the first burn.

### #17

Posted 06 March 2017 - 23:40

It doesn't work like that with rockets. The expelled mass must be substantial for the rocket engine to work, it's not a pure jet engine or a turbofan. It's a collision ran backwards.

And to refuel the rocket you'd have had to accelerate the fuel and oxidizer.

Just look up the weights in a rocket. Or the Shuttle on the pad:

The Space Shuttle weighed 165,000 pounds empty. Its external tank weighed 78,100 pounds empty and its two solid rocket boosters weighed 185,000 pounds empty each. Each solid rocket booster held 1.1 million pounds of fuel. The external tank held 143,000 gallons of liquid oxygen (1,359,000 pounds) and 383,000 gallons of liquid hydrogen (226,000 pounds). The fuel weighed almost 20 times more than the Shuttle. At launch, the Shuttle, external tank, solid rocket boosters and all the fuel combined had a total weight of 4.4 million pounds. The Shuttle could also carry a 65,000 payload.

**Edited by saudoso, 06 March 2017 - 23:47.**

### #18

Posted 07 March 2017 - 01:15

If the frame of reference is moving the same still applies. Suppose it is moving at 1000 m/s. The small body is accelerate from 0 to 10 m/s in that frame of reference.

In universal frame of reference the body has accelerated from 1000 to 1010. It's KE has increased from 1/2*1*1000^2=500kJ to 1/2*1*1010^2= 510050J, so it's gone up by 10050 instead of 50! But the work done is 5N*average speed over the 2 seconds, 1005, *2 seconds=10050. Exactly the same number. In other words frames of reference that are moving at constant speed make no odds to the energy balance.

### #19

Posted 07 March 2017 - 03:58

It doesn't work like that with rockets. The expelled mass must be substantial for the rocket engine to work

If a 1,000 ton rocket expels 1 kg from its exhaust at 1,000 m/s the velocity of the rocket will increase by 0.001 m/s and the mass of the rocket will reduce to 999.999 ton. The next kg expelled will increase the velocity by a similar amount.

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### #20

Posted 07 March 2017 - 07:26

If a 1,000 ton rocket expels 1 kg from its exhaust at 1,000 m/s the velocity of the rocket will increase by 0.001 m/s and the mass of the rocket will reduce to 999.999 ton. The next kg expelled will increase the velocity by a similar amount.

Say the rocket engine is 100% efficient.

It expels a quantity x of fuel and accelerates to 10m/s. It does a second burn of x fuel and accelerates to 20m/s.

It now has 4 times more energy than after the first burn but yet has only burned the same amount of fuel as before. Where is all this extra energy coming from if E=.5.m.v^2.

Please treat the rocket has having the same mass throughout.

### #21

Posted 07 March 2017 - 08:13

My old rusted gears are rusting.

It's obvious there is something wrong with this, but I'll think a bit longer before I commit again.

### #22

Posted 07 March 2017 - 09:11

### #23

Posted 07 March 2017 - 09:12

### #24

Posted 07 March 2017 - 09:24

Ali_g look at my example again. The speed of the reference frame is irrelevant.

Yup, I see it now. Just forget about the energy liberated by the chemical reaction and consider the mechanics of it.

10N of thrust at 10m/s = 100W

10N of thrust at 20m/s = 200W

Even if the burn is exactly the same.

### #25

Posted 08 March 2017 - 03:22

Even "simple physics" or "high school physics" is not really simple. It is easy enough to work with the various equations derived from Newton's Laws (he really was a clever bastard) and quote them endlessly as "explanations" - but can you explain to somebody the difference in commonsense terms between momentum and kinetic energy without using any equations?

Agreed. I understood the mathematics of this question for 20 years, but I still can't grasp this intuitively even though I'm generally pretty good at understanding things intuitively. Probably part of the reason is that it's hard to grasp polynomial physical relationships that don't involve areas or volumes, it just feels unnatural. And forget about trying to understand why 10-20 requires four times as much energy as 0-10 regardless of how fast the earth on which all this is done is moving and rotating.

### #26

Posted 09 March 2017 - 06:34

Say the rocket engine is 100% efficient.

It expels a quantity x of fuel and accelerates to 10m/s. It does a second burn of x fuel and accelerates to 20m/s.

It now has 4 times more energy than after the first burn but yet has only burned the same amount of fuel as before. Where is all this extra energy coming from if E=.5.m.v^2.

Please treat the rocket has having the same mass throughout.

OK. You are considering the initial state of the rocket as your frame of reference. What you find is, that in this frame of reference, the kinetic energy of the system after the second burn is twice that of the first burn. The rocket itself gains three times as much energy during the second burn but the energy of the reaction mass (rocket exhaust) is less for the second burn - by the same amount. (BTW KE of the exhaust >> KE of the rocket)

### #27

Posted 09 March 2017 - 13:04

I still think Greg nailed it in post 2, and it boils down to the difference between work and force.

Burning the same amount of fuel would lead to the same thrust (force) (all other forces being neglected), but the work done is significantly different. Thus,

Change in Total Energy= Change in Kinetic+Change in Potential Energy = Work Done **over** the System (hence the positive sign)

**Edited by turin, 09 March 2017 - 15:37.**

### #28

Posted 09 March 2017 - 15:51

This should be a simple question.

So, kinetic energy has the formula 1/2.m.v^2

So, what about the following scenario. An object weighing a kg is accelerated to 10m/s. In the above formula, this will take 50 joules to achieve.

If the same object is accelerated from rest to 20m/s it would take 200 joules.

Therefore, if the same object is accelerated from 10m/s to 20m/s, shouldn't it take 150 joules to do so? Why does going from to 10 to 20 take so much more energy or am I misunderstanding the formula?

I believe it depends on the reference frame.

### #29

Posted 09 March 2017 - 20:28

It has nothing to do with the reference frame. You have to understand how algebra works.

If E1 = 0.5m*V1^2 for the energy to accelerate mass "m" from rest to velocity "V1" and E2 = 0.5m*V2^2 is the energy to accelerate mass "m" from rest to V2, then the energy required to accelerate mass "m" from V1 to V2 is E2-E1 = 0.5m*V2^2 - 0.5m*V1^2 = 0.5m*(V2^2 - V1^2)

The fact that kinetic energy depends on the square of the velocity is why going from 0 to 10 mps takes less energy than going from 10 mps to 20 mps, even though the velocity change is the same 10 mps.

### #30

Posted 10 March 2017 - 10:02

The problem being: it doesn't in that case.

If you take the burned fuel necessary to do the deed, there's absolutely no difference. A rocket engine will burn a steady rate of fuel, expel a constant mass of exhaust and produce constant thrust. Nothing changes. At all. NADA.

Kinetic energy is relative to the observer and will vary depending on relative observer speeds, bearing no effect at all on the local characteristics of the observed object.

### #31

Posted 11 March 2017 - 14:05

The problem being: it doesn't in that case.

If you take the burned fuel necessary to do the deed, there's absolutely no difference. A rocket engine will burn a steady rate of fuel, expel a constant mass of exhaust and produce constant thrust. Nothing changes. At all. NADA.

Kinetic energy is relative to the observer and will vary depending on relative observer speeds, bearing no effect at all on the local characteristics of the observed object.

No. It is because thrust is just the force, which is indeed constant in this case (and so is acceleration, by Newton's 2nd).

To change KE you need to do **work **., which is the force (thrust) times distance.

As Greg pointed out in post 2, the distance it travels when going from 0-10 m/s **is different** than when it goes from 10-20 m/s. Hence you need to provide more energy when going from 10-20 than when going from 10-20.

**Edited by turin, 11 March 2017 - 19:03.**

### #32

Posted 11 March 2017 - 22:53

The apparent paradox in the case of the rocket is this. In going from 0 - 10 m/s the kinetic energy of the rocket increases by m/2 x 100 and the fuel burn is X. In going from 10 - 20 the kinetic energy increases by m/2 x (400 - 100) = m/2 x 300 ie 3 times as much, yet the fuel burn is once again X. Where did all the extra energy come from? The answer is in the exhaust. The kinetic energy of the exhaust from the first burn is higher than that of the second burn (m/2 x 200 joules higher).

### #33

Posted 11 March 2017 - 23:41

To change KE you need to do

work., which is the force (thrust) times distance.

The measure of work you get is relative to frame of reference. As are the absolute values of kinetic and potential energy. Demonstrated by our guru Greg on post 18.

As Greg pointed out in post 2, the distance it travels when going from 0-10 m/s

is differentthan when it goes from 10-20 m/s. Hence you need to provide more energy when going from 10-20 than when going from 10-20.

Wrong. The energy within the rocket is enough to keep acceleration. It will burn exactly the same amount of fuel to go from 0 to 10m/s and 10 to 20m/s - in vacuum, that is.

### #34

Posted 11 March 2017 - 23:46

Actually even in the rocket example, the work done on the rocket from 10-20 is three times the work done on the rocket from 0-10. The work done on the rocket ** plus** the work done on the exhaust gas is the same for both burns.

### #35

Posted 12 March 2017 - 10:10

Which brings us to the point of selecting a correct system boundary.

### #36

Posted 12 March 2017 - 23:24

Wrong. The energy within the rocket is enough to keep acceleration. It will burn exactly the same amount of fuel to go from 0 to 10m/s and 10 to 20m/s - in vacuum, that is.

Again you are mixing energy with force. You can have energy (say kinetic) and have no acceleration, as when an object moves at a constant speed. Having energy does not mean you can keep accelerating. To accelerate, you need a net force which must do work. Even when you think in terms of mechanical energy of a conservative system (like a frictionless pendulum), the change in kinetic energy is due to the work of gravity, which can be rearranged as potential energy.

You should get the same answer regardless of the system of reference (c.f. post 18), unless your reference frame is non inertial (ie, it is accelerating), in which case there is an extra "fictitious" force. Post 18 has a small disclaimer towards then when he says "moving at constant speed" (inertial frame of reference)

### #37

Posted 12 March 2017 - 23:50

I meant chemical energy available to be released by the rocked by combustion. No matter how fast the rocked seems to you, if it took one Mj of energy energy consumed by the engine for it to accelerate from 0 to 10, it will take exactly the same 10Mj for it to be acelerated from 10 to 20, being the total rocket mass considered fixed, of course. Contrary to your statement below:

As Greg pointed out in post 2, the distance it travels when going from 0-10 m/s

is differentthan when it goes from 10-20 m/s.Hence you need to provide more energy when going from 10-20 than when going from 10-20.

When you say provide, I understand consume energy from a material source, not calculated from a fast moving frame of reference. That amount of materially consumed energy stays the same. Was I clear now?

GG nailed it on post 34 when he said you must consider the energy of exhaust gas. If you take that in account, both burns add exactly the same amount of kinetic energy to the system.

### #38

Posted 12 March 2017 - 23:51

It does take a while to nut out. For example if the speed of the frame of reference is r, and the initial speed relative to it is u, and final velocity t seconds later is v

it is not immediately obvious that f*(u+v)/2*t=1/2*m*(v^2-u^2) is in fact identical to f*((u+v)/2+r)*t=1/2*m*((v+r)^2-(u+r)^2)

but if you grind through the algebra (it's about 10 lines, and the initial expansion is unwieldy) it is in fact the case.

### #39

Posted 13 March 2017 - 01:32

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### #40

Posted 13 March 2017 - 12:39

It does take a while to nut out. For example if the speed of the frame of reference is r, and the initial speed relative to it is u, and final velocity t seconds later is v

it is not immediately obvious that f*(u+v)/2*t=1/2*m*(v^2-u^2) is in fact identical to f*((u+v)/2+r)*t=1/2*m*((v+r)^2-(u+r)^2)

but if you grind through the algebra (it's about 10 lines, and the initial expansion is unwieldy) it is in fact the case.

Are you saying that

v^2-u^2 = (v+r)^2-(u+r)^2 or did you mean something else?

### #41

Posted 13 March 2017 - 18:23

A lot of this is beyond my maths but one thing relevant to rockets is the way they tilt over after a vertical take off. I used to think this was just lining up to go round the world ( being dumb) but ts not.

It is a deliberate strategy to reduce the gravitational drag whilst speed builds up.. As you lean the rocket over on its side more thrust can be devoted to acceleration which gives more centrifugal force to offset gravity.

If it just went straight up the fuel would run out quicker versus altitude.

### #42

Posted 13 March 2017 - 20:15

I meant chemical energy available to be released by the rocked by combustion. No matter how fast the rocked seems to you, if it took one Mj of energy energy consumed by the engine for it to accelerate from 0 to 10, it will take exactly the same 10Mj for it to be acelerated from 10 to 20, being the total rocket mass considered fixed, of course. Contrary to your statement below:

When you say provide, I understand consume energy from a material source, not calculated from a fast moving frame of reference. That amount of materially consumed energy stays the same. Was I clear now?

GG nailed it on post 34 when he said you must consider the energy of exhaust gas. If you take that in account, both burns add exactly the same amount of kinetic energy to the system.

So we were talking of different things then. I have been referring to the mechanical energy (potential plus kinetic) of the rocket as a whole, which I think was the point of the OP. What you are alluding to is potential energy of the gas or fuel, that is transformed into kinetic energy, heat and work by the thermodynamics (most notably the first law). That kinetic energy of the gas, in fact, its momentum flux yields a net force as it leaves the control volume of choice. That force is constant and is the one I am alluding to.

### #43

Posted 13 March 2017 - 21:46

I learned all this stuff in portuguese more than 3 decades ago, so I'm prone to mess the terminology up

### #44

Posted 13 March 2017 - 21:54

A lot of this is beyond my maths but one thing relevant to rockets is the way they tilt over after a vertical take off. I used to think this was just lining up to go round the world ( being dumb) but ts not.

It is a deliberate strategy to reduce the gravitational drag whilst speed builds up.. As you lean the rocket over on its side more thrust can be devoted to acceleration which gives more centrifugal force to offset gravity.

If it just went straight up the fuel would run out quicker versus altitude.

Actually I think it is lining itself up to go into orbit.

### #45

Posted 13 March 2017 - 23:12

I learned all this stuff in portuguese more than 3 decades ago, so I'm prone to mess the terminology up

### #46

Posted 14 March 2017 - 04:06

The equation with r in it reduces to the one without r in it.

**Edited by Greg Locock, 14 March 2017 - 04:24.**

### #47

Posted 14 March 2017 - 04:57

The odd symbols on the keyboard make the equations slightly hard to understand - do you mean v squared = u squared +2as?

### #48

Posted 14 March 2017 - 06:16

v > u

(v+r)^2-(u+r)^2

v^2 + 2rv + r^2 - u^2 - 2ru - r^2

v^2 - u^2 + 2r(v-u)

The last bit is always greater than zero with the constraints given.

### #49

Posted 14 March 2017 - 10:03

5/10 at best Now look at the LHS of the equations, you need to sort those out.

### #50

Posted 15 March 2017 - 15:18

If it just went straight up the fuel would run out quicker versus altitude.

When the rocket takes off, there is a huge force pushing it up. No matter how clever the engineers, the lift force will not be equal over 360 degrees of the rocket aperture. So the rocket's departure from the launch site will be a bit wobbly. After a few seconds, the control systems in the rocket have enough data about the engines to take control and aim for the target.