| Topic: |
Science > Physics |
| User: |
"numberdude" |
| Date: |
04 Aug 2004 09:24:03 PM |
| Object: |
Getting something for nothing? |
I was pondering the question, "Does it take the same amount of energy to
accelerate a mass from 10 to 20 mph (in say, one second) as it does to
accelerate it from 30 to 40mph (also in one second)?" In other words, for a
change in 10mph per second, will it always take the same amount of energy?
Kinetic energy = 1/2mv^2, so the change in KE = 1/2m(v2^2 - v1^2 )
For the case where v2 = 20mph and v1 = 10mph
Change in KE = 1/2m(400 - 100) = 150m
For the case where v2 = 40mph and v1 = 30mph
KE = 1/2m(1600 - 900) = 350m
It takes more than TWICE as much energy in the second case!
Now here's the puzzle: It takes a certain amount of force to produce the
change in velocity in both cases. Since the changes in velocity took place
in the same amount of time in both cases, the accelerations are the same.
Now we have equal masses moving with equal accelerations (but at different
velocities). Since the force needed to produce the acceleration is, F = ma,
the forces must be the same in both cases. But this means that equal forces
acting for equal times on equal masses will produce different changes in
kinetic energy at different velocities! It seems that at higher initial
velocities a given force will produce a greater change in kinetic energy.
But this is like getting something for nothing, which I know shouldn't
happen. Am I missing something?
Allen
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| User: "Old Man" |
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| Title: Re: Getting something for nothing? |
05 Aug 2004 11:49:12 PM |
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"numberdude" <numberdude@netzero.net> wrote in message
news:0JgQc.45817$vX4.15912@cyclops.nntpserver.com...
I was pondering the question, "Does it take the same amount of energy to
accelerate a mass from 10 to 20 mph (in say, one second) as it does to
accelerate it from 30 to 40mph (also in one second)?" In other words, for
a
change in 10mph per second, will it always take the same amount of energy?
Kinetic energy = 1/2mv^2, so the change in KE = 1/2m(v2^2 - v1^2 )
For the case where v2 = 20mph and v1 = 10mph
Change in KE = 1/2m(400 - 100) = 150m
For the case where v2 = 40mph and v1 = 30mph
KE = 1/2m(1600 - 900) = 350m
It takes more than TWICE as much energy in the second case!
delta_KE = ( m / 2 ) ( v1 + v2 ) delta_v
for a given delta_v, the delta_KE is proportional to the sum
of the initial and final speeds.
Now here's the puzzle: It takes a certain amount of force to produce the
change in velocity in both cases. Since the changes in velocity took place
in the same amount of time in both cases, the accelerations are the same.
But, for uniform acceleration and equal time interval, the
distance traveled, delta_x, is proportional to the average
speed:
delta_x = (1 / 2) (v1 + V2) delta_t
and the work is given by
Work = Force*delta_x = (Force / 2) (v1 + V2) delta_t
Thus, the Work is proportional to the sum of the initial
and final speeds. As a check, we set Work = delta_KE :
(Force / 2) (v1 + V2) delta_t = ( m / 2 ) ( v1 + v2 ) delta_v
or Force = m ( delta_v / delta_t )
Which, for constant delta_v and constant delta_t, is also
a constant of the correct magnitude. No free lunch.
[Old Mam]
Now we have equal masses moving with equal accelerations (but at different
velocities). Since the force needed to produce the acceleration is, F =
ma,
the forces must be the same in both cases. But this means that equal
forces
acting for equal times on equal masses will produce different changes in
kinetic energy at different velocities! It seems that at higher initial
velocities a given force will produce a greater change in kinetic energy.
But this is like getting something for nothing, which I know shouldn't
happen. Am I missing something?
Allen
.
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| User: "" |
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| Title: Re: Getting something for nothing? |
06 Aug 2004 09:03:17 AM |
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"numberdude" <numberdude@netzero.net> wrote in message news:<0JgQc.45817$vX4.15912@cyclops.nntpserver.com>...
[old chestnut involving W = F.d (not F.t) snipped]
Now you have to answer one. No fair searching back through
the sci.physics archives on google, as this one has been
beaten to death dozens of times.
Consider a "rocketship" consisting of a 100 kg mass
with a spring that can launch a 1 kg mass at 100 m/s.
If the ship is at rest relative to the lab, and on a frictionless
surface (one of those air hockey tables will do) then it of
course winds up going 1 m/s. And it gains 1/2mv^2 or 50 Joules
of energy.
But what if it starts going 1 m/s, and fires the same mass backwards.
It then must finish up going 2 m/s, for a final kinetic energy of
200 Joules, or a gain of 150 Joules, three times as much.
How about that numberdude? It looks like the same question you
posed, but how can it be the F.d vs F.t thing? The same exact
spring is doing the same exact thing both times.
After you figure that one out, do it for arbitrary masses and
arbitrary starting velocities. Then when that's too easy for you,
do it accounting for special relativity. Once you get that far,
you're almost ready to do particle physics.
Socks
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| User: "Bjoern Feuerbacher" |
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| Title: Re: Getting something for nothing? |
05 Aug 2004 03:21:03 AM |
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numberdude wrote:
I was pondering the question, "Does it take the same amount of energy to
accelerate a mass from 10 to 20 mph (in say, one second) as it does to
accelerate it from 30 to 40mph (also in one second)?" In other words, for a
change in 10mph per second, will it always take the same amount of energy?
Kinetic energy = 1/2mv^2, so the change in KE = 1/2m(v2^2 - v1^2 )
For the case where v2 = 20mph and v1 = 10mph
Change in KE = 1/2m(400 - 100) = 150m
For the case where v2 = 40mph and v1 = 30mph
KE = 1/2m(1600 - 900) = 350m
It takes more than TWICE as much energy in the second case!
Now here's the puzzle: It takes a certain amount of force to produce the
change in velocity in both cases. Since the changes in velocity took place
in the same amount of time in both cases, the accelerations are the same.
Now we have equal masses moving with equal accelerations (but at different
velocities). Since the force needed to produce the acceleration is, F = ma,
the forces must be the same in both cases. But this means that equal forces
acting for equal times on equal masses will produce different changes in
kinetic energy at different velocities!
Completely right. This is due to the simple reason that work is
force times *distance*, not times time (actually, the integral of the
force along the path on which the object on which it acts moves). In the
second case, the distance is greater, hence the work is greater, hence
the change in energy is greater.
It seems that at higher initial
velocities a given force will produce a greater change in kinetic energy.
A given force acting for the same time in the same direction, yes.
But this is like getting something for nothing,
No, not at all. Why do you think so?
which I know shouldn't happen. Am I missing something?
Yes. W=F*s, not W=F*t
Bye,
Bjoern
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