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Re: Repeal of the National Speed Limit Law
In article <95187.072906MICHAEL@MAINE.MAINE.EDU>, I wrote:
>
> W = F * D (1)
>
>Work is force times distance. Pretty basic comic book physics.
>
> E = W / T (2)
>
>Energy is work over time. Again, basic comic book physics. If we assign
>units to the values on the right (F in Newtons, D in Meters, T in
>seconds) we have:
>
> Energy = Newtons * Meters / Second (3)
Having consulted my physics text, I can now say that I mixed up my
terms. Energy and Work are the same thing. What I called energy is
actually power. So my terms are a little skewed, but that doesn't
invalidate my proof, which is based on F=MA.
So, while you can expend varying amounts of power accelerating from 0-60
mph, the energy used is the same. According to said physics text, the
magnitute of energy used to effect a velocity change is ALWAYS equal to the
change in kinetic energy (relative to a frame of reference, of course).
Later on, I wrote:
>The energy used to effect this velocity change can be calculated using:
>
> E = F * Vd
This is not quite right. The equation is actually:
P = (F * D) / T
The distance traveled is of course relative to a frame of reference. This
would be given by the average relative velocity times the acceleration
interval. For an observer in a vehicle moving at X-10 mph, this would be:
P = (5000 kg ft/sec/sec * (7.5 ft/sec * 3 sec) / 3 sec = 375000 power units
The work would then be given by:
2 2
E = 112500 kg ft / sec
Or:
2
E = 112500 kg (ft/sec)
Which if we plug in the original mass and velocity change numbers, give us
1 2 2
E = --- * (1000 kg) * (15 ft/sec) = 112500 kg (ft/sec)
2
Which matches the kinetic energy equation:
1 2
KE = --- M V
2
I then wrote:
>So obviously something is missing in the kinetic energy picture. I'm not
>going to try to guess what it is, I'd rather go dig out my physics text,
>which is at home. It may be that Jeff and Lloyd are applying the KE equation
>in ways that don't apply (gee, do you think so?).
So now we can examine what is wrong with this picture. Let's do a little
more math. Since the KE change is supposed to match the energy expended
(that is, the work done) in changing velocity, how can the time always
be the same for a given velocity change?
Let's examine our scenario of a 75-65 mph change over 3 seconds for a
car massing 1000 Kg. The beginning KE with respect to a stationary
observer is:
2
KE0 = 0.5 * 1000Kg * (110 ft/sec) = 6050000 energy units
And the ending KE is:
2
KE1 = 0.5 * 1000Kg * (95 ft/sec) = 4512500 energy units
For a change in KE of 1537500 energy units. Those same energy units
can be expressed as force times distance. We calculated the force to
be 5000 units of force. So now we have to figure out how much distance
was involved. We can average the velocity of the car over the interval
to give 102.5 ft/sec times 3 seconds for 307.5 feet. So the energy
should be given by:
KEd = 307.5 ft * 5000 force units = 1537500 energy units
Which matches perfectly. We can do the same calculation for a change from
65 mph to 55 mph:
2
KE0 = 0.5 * 1000 Kg * (95 ft/sec) = 4512500 energy units
2
KE1 = 0.5 * 1000 Kg * (80 ft/sec) = 3200000 energy units
For a difference of 1312500 which is less than for the change from 75
to 65 mph, as expected. Now, using the work equation, the average
velocity of the car over the 3 seconds is 87.5 ft/sec, giving a
distance traveled of 262.5 feet, so the KE change should be:
KEd = 262.5 ft * 5000 force units = 1312500 energy units
Which again matches. So as you can see, the time is always the same,
even though the kinetic energy change is geometrically increasing.
My terminology in describing the energy exchange was off, which is why
my units didn't always match up. It's been a while since I studied
energy. Nevertheless, my physics based on F=MA were correct and once
I found the flaw in my energy calculations, were verified based on
kinetic energy change.
Michael Johnson, Relay Technology, Inc.
michael@maine.maine.edu, michaelj@relay.relay.com
References: