I was pondering the question...

Hi,

I was pondering the question, "Does it take more energy to accelerate a mass
from say, 20mph to 30mph, or from 30mph to 40mph?"

Here's the math:

KE = 1/2(mv^2)

For the case where v1 = 30mph and v2 = 20mph:

Change in KE = 1/2*m(900 - 400) = 250m

For the case where v1 = 40mph and v2 = 30mph:

Change in KE = 1/2*m(1600 - 900) = 350m

So it takes more energy in the second case. Now it takes some force to
produce the accelerations. You'd think (at least I thought) it would take
more force in the second case, but no, the forces are the same. Since F = ma
and the masses and accelerations are the same in both cases, the forces must
be the same. It seems like we're getting something for nothing, which I
believe cannot be. So, what am I missing (besides a PhD in physics)?

Allen

P.S. Sorry if this posts twice. I posted yesterday but didn't see it here.

>
You may want to remember that the energy you put in is force times distance.
Calculate the distance it takes in each case for the change in energy, at
constant force, or what the force would be, for constant distance. It'll become
clear to you then.

On Thu, 5 Aug 2004 21:07:23 -0400, "numberdude"
<[Only registered users see links. ]> wrote:

That should be (delta v squared), not (v1 squared - v2 squared).
Neglecting the difference in rolling friction and air drag the
energy required would be the same.

actually there is a difference. Reletivity theory makes energy a little larger
than linear when velocity is faster. It is very small though.

Dear numberdude:

"numberdude" <[Only registered users see links. ]> wrote in message
news:XGAQc.65287$[Only registered users see links. ]. ..
mass
ma
must

take the derivative...
dKE/dt = m(v . dv/dt).
note that...
F=ma=m*dv/dt
therefore...
dKE/dt = v . a

The product of velocity and force is what you could look at. You have to
work a lot harder to apply the force to a moving "energy sink"... for each
second. What you haven't looked at is how long it took to get from "20mph
to 30mph" as compared to "30mph to 40mph".

Maybe this will help. Maybe I screwed up...

David A. Smith

"KellyClarksonTV" <[Only registered users see links. ]> wrote in message
news:[Only registered users see links. ].cs.com...
larger

this is true, although a much more marked difference would be the effect of
air resistance, which gets greater the faster you go, so increases in speed
from higher speeds would require greater energy input if performed in the
real world

Dear Jeremy Watts:

"Jeremy Watts" <[Only registered users see links. ]> wrote in message
news:Y6HQc.18$[Only registered users see links. ].net...
of
speed

The OPs math was correct in a vacuum. It does take more energy to go from
20 mph to 30 mph, than it does from 10 mph to 20 mph. Consider braking
distances, which are equivalent to constant forces...

The "air resistance" thing is confusing you.

David A. Smith

"numberdude" <[Only registered users see links. ]> wrote in message news:<XGAQc.65287$[Only registered users see links. ] >...

Also, if you apply the same force to something moving backwards:

v1 = - 20mph
v2 = - 30mph

Change in KE = 1/2*m(400 - 900) = - 250m

If you apply the same force to make an object go slower, you can get
energy out of it. Regenerative braking in hybrid autos would be a
good example.

The key thing to remember here is the difference between force and
power (the rate at which energy is used). A simple table can exert an
upward force on an object, to support it against gravity, for years
without tiring. You only need energy if the object is moving in the
direction you are applying the force; power = force * velocity. For
example, to lift the weight upward, you would need to consume energy.
And to accelerate a mass that's moving forward, you need to use
energy, and you'd need it at a faster rate (more power) the faster the
mass is moving.

Guys,

I don't want to sound like a jerk, but noone answered my question. First of
all, is there anything wrong in my reasoning? Did I calculate the change in
KE correctly? Am I right in concluding that the force is the same in both
cases? I think I did this correctly. If the force really is the same, then
isn't that surprizing? It is to me. Also, forget air resistance - this is
an idealized situation to get to the heart of the matter. Make believe it's
in a vacuum for simplicity. Also, I mentioned that the times are the same in
both cases. It seems to me that the only equations necessary are KE =
(1/2)mv^2 and F=ma.

I welcome all comments.

Thanks,
Allen

"numberdude" <[Only registered users see links. ]> wrote in message
news:XGAQc.65287$[Only registered users see links. ]. ..
mass
ma
must

Your calculations are correct; it *does* take more energy to accelerate from
30 mph to 40 mph than it does to acclerate a mass from 20 mph to 30 mph.

ma
must

Actually, the forces *don't* enter into the problem as you have stated it.

Acceleration is the rate of change in velocity per unit time, but you have
only specified the change in velocity.

In the language of differential calculus,
a = dv/dt

In the language of algebra,
a = (v2 - v2)/(t2 - t1)

The question of "How much time does it take to accomplish this
acceleration?" is left wide open.

A very small force can accomplish a large acceleration if it is given
proportionately more time to do it.

It would take Nolan Ryan a lot less time to accelerate a baseball from 30 to
40 mph than it would take a 6-year old child, but both could do it, and the
energies would be the same regardless of how much time it takes to put that
energy into the baseball.

HTH


Tom Davidson
Richmond, VA