Change in kinetic energy of an accelerating object
in [Physics]
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From: boltar2003 on 26 May 2010 10:16 Before anyone asks , no this isn't some school project I'm working on. Its a simple problem I can't figure out for myself. If you have a rocket motor giving a constant thrust then the rocket - not accounting for weight change due to fuel use - will accelerate at a constant rate. However if say its accelerating at 1m/s/s then it will take twice as long to accelerate from 2 m/s to 4 m/s as it did to get from 1 m/s to 2 m/s. Yet going by 0.5 mv^2 the kinetic energy for a rocket weighing 1 kilo at the given speeds is: 1 m/s = 0.5J 2 m/s = 2J 4 m/s = 8J So in the first second it gained 1.5J but in the next 2 seconds it gained 6J. I don't get it? The motor wasn't putting out any extra energy so where did it come from? B2003
From: Androcles on 26 May 2010 11:35 <boltar2003(a)boltar.world> wrote in message news:htjaf3$sqb$1(a)speranza.aioe.org... | Before anyone asks , no this isn't some school project I'm working on. Its | a simple problem I can't figure out for myself. | | If you have a rocket motor giving a constant thrust then the rocket - not | accounting for weight change due to fuel use - will accelerate at a constant | rate. However if say its accelerating at 1m/s/s then it will take twice as | long to accelerate from 2 m/s to 4 m/s as it did to get from 1 m/s to 2 m/s. t = 0, v = 0, a= 1, height =1/2 at^2 = 0 t = 1, v = 1, h = 0.5 t = 2, v = 1+1 = 2, h = 2 t = 3, v = 2+1 = 3, h = 4.5 t = 4, v = 3+1 = 4, h = 8 | Yet going by 0.5 mv^2 the kinetic energy for a rocket weighing 1 kilo at the | given speeds is: | | 1 m/s = 0.5J | 2 m/s = 2J | 4 m/s = 8J | | So in the first second it gained 1.5J but in the next 2 seconds it gained | 6J. I don't get it? The motor wasn't putting out any extra energy so where | did it come from? | In the first second it gained 1/2 m(v1-v0)^2 = 1/2 * 1 * 1^2 = 0.5J, which is also numerically the height (for m = 1).
From: boltar2003 on 26 May 2010 11:46 On Wed, 26 May 2010 16:35:11 +0100 "Androcles" <Headmaster(a)Hogwarts.physics_z> wrote: ><boltar2003(a)boltar.world> wrote in message >news:htjaf3$sqb$1(a)speranza.aioe.org... >| Before anyone asks , no this isn't some school project I'm working on. Its >| a simple problem I can't figure out for myself. >| >| If you have a rocket motor giving a constant thrust then the rocket - not >| accounting for weight change due to fuel use - will accelerate at a >constant >| rate. However if say its accelerating at 1m/s/s then it will take twice as >| long to accelerate from 2 m/s to 4 m/s as it did to get from 1 m/s to 2 >m/s. > >t = 0, v = 0, a= 1, height =1/2 at^2 = 0 >t = 1, v = 1, h = 0.5 >t = 2, v = 1+1 = 2, h = 2 >t = 3, v = 2+1 = 3, h = 4.5 >t = 4, v = 3+1 = 4, h = 8 Thats all very well, but where is the energy coming from? The engine is only providing a constant amount of power whereas the kinetic energy increase is non linear and so when the difference in kinetic energy between 1 second and the next reaches the amount of joules/sec the engine is providing then surely the rockets acceleration should slow down and tail off? But the engine is also providing a constant force which means the acceleration should be constant until the engine stops. So we have a contradiction. B2003
From: PD on 26 May 2010 12:15 On May 26, 9:16 am, boltar2...(a)boltar.world wrote: > Before anyone asks , no this isn't some school project I'm working on. Its > a simple problem I can't figure out for myself. > > If you have a rocket motor giving a constant thrust then the rocket - not > accounting for weight change due to fuel use - will accelerate at a constant > rate. However if say its accelerating at 1m/s/s then it will take twice as > long to accelerate from 2 m/s to 4 m/s as it did to get from 1 m/s to 2 m/s. > Yet going by 0.5 mv^2 the kinetic energy for a rocket weighing 1 kilo at the > given speeds is: > > 1 m/s = 0.5J > 2 m/s = 2J > 4 m/s = 8J > > So in the first second it gained 1.5J but in the next 2 seconds it gained > 6J. I don't get it? The motor wasn't putting out any extra energy so where > did it come from? > > B2003 The kinetic energy comes from work. The difficulty you have is thinking that with a constant force, there should be a constant amount of work. But this is not the case. Work is force x distance. What this means is that if you have a force of 20 N acting on a body that goes 10 m in one second, it will contribute a different amount of energy than the same force of 20 N acting on a body that goes 30 m in the same second. The same applies not only to a rocket but to a melon dropped under the force of gravity. The force of gravity stays the same throughout the drop, but the kinetic energy increase (that is, the work) is three times more in the second second than it is in the first second, because it falls three times as far in that next second. But going back to the rocket example, your next question is likely to be, "But the ultimate source of the kinetic energy is the chemical potential energy in the fuel that is being burned, and that's being burned at a constant rate, releasing chemical energy at a constant rate." This is true, and if you account for the kinetic energy of both the rocket AND the exhaust, then you will see that none is lost -- it's only the share between the exhausted fuel and the rocket that changes.
From: JimboCat on 26 May 2010 12:16
On May 26, 10:16 am, boltar2...(a)boltar.world wrote: > Before anyone asks , no this isn't some school project I'm working on. Its > a simple problem I can't figure out for myself. > > If you have a rocket motor giving a constant thrust then the rocket - not > accounting for weight change due to fuel use - will accelerate at a constant > rate. However if say its accelerating at 1m/s/s then it will take twice as > long to accelerate from 2 m/s to 4 m/s as it did to get from 1 m/s to 2 m/s. > Yet going by 0.5 mv^2 the kinetic energy for a rocket weighing 1 kilo at the > given speeds is: > > 1 m/s = 0.5J > 2 m/s = 2J > 4 m/s = 8J > > So in the first second it gained 1.5J but in the next 2 seconds it gained > 6J. I don't get it? The motor wasn't putting out any extra energy so where > did it come from? Kinetic energy is frame-dependent (because velocity is frame- dependent), and the rocket is continuously changing frames (more accurately, it isn't in an inertial frame at all). So the energy comparison is simply invalid: it doesn't mean anything. Jim Deutch (JimboCat) -- "the content of physics doesn't need the language of causality. I think it is analogous to the way evolutionary biologists talk about teleological arguments. Teleology is just shorthand for the way natural selection works. Causality is just shorthand for the way coupled systems behave." [Bernard B. Beard] |