Question about energy requirements for acceleration

[halloweenjack]

I read the following on howstuffworks.com, and their explanation just doesn't seem correct to me.

"One of the interesting things about kinetic energy is that it increases with the velocity squared. This means that if a car is going twice as fast, it has four times the energy. You may have noticed that your car accelerates much faster from 0 mph to 20 mph than it does from 40 mph to 60 mph. Let's compare how much kinetic energy is required at each of these speeds. At first glance, you might say that in each case, the car is increasing its speed by 20 mph, and so the energy required for each increase must be the same. But this is not so.

We can calculate the kinetic energy required to go from 0 mph to 20 mph by calculating the KE at 20 mph and then subtracting the KE at 0 mph from that number. In this case, it would be 1/2*m*20^2 - 1/2*m*0^2. Because the second part of the equation is 0, the KE = 1/2*m*20^2, or 200 m. For the car going from 40 mph to 60 mph, the KE = 1/2*m*602 - 1/2*m*40^2; so KE = 1,800 m - 800 m, or 1000 m. Comparing the two results, we can see that it takes a KE of 1,000 m to go from 40 mph to 60 mph, whereas it only takes 200 m to go from 0 mph to 20 mph. "

The way they explain this implies that it is more difficult for an object that is already in motion to accelerate further. Wouldn't this imply that a greater force is necessary to accelerate an object at higher speeds than at lower speeds? As far as I can remember F=ma and neither acceleration nor force is affected by current velocity (excluding velocities that are near the speed of light, of course).

Also, doesn't kinetic energy depend on frame of reference as well? If you are already moving at 20 mph as in the example at a constant speed, couldn't your frame of reference just as easily be 0 mph? I realize that with the way a car's engine works and the way it is geared, air resistance, friction, etc, this may not be true, but what about if it was a spaceship in a perfect vacuum? It doesn't seem that if you had two identical rockets used for thrust that the first rocket used would accelerate you more or more quickly than the second rocket used does just because your velocity relative to Earth was now higher after firing off the first rocket.

Thank you for your explanations.

 

[DeShark]

I agree completely with you. Maybe you should write to them and explain things a little better. The energy of an object due to its motion is found from the definitions:

\vec{F}=m\vec{a} and E = \int{\vec{F}\cdot d \vec{x}}

From these, the Kinetic energy just falls out with a little calculus. It's nice to have Energies, cause they're scalar and they're generally nice to work with. Forces cause acceleration, which is observable. It's good to be able to define potentials for conservative force fields, because then the Force is the derivative (or gradient) or the scalar field. Energy conservation is an observed fact, and having the kinetic energy as it is (proportional to mass and the velocity squared) just works in all the observations. It simplifies a hell of a lot of problems, which is what physics is about, no?

 

[halloweenjack]

I'd like to write them but I'm not really qualified, I just took a couple classes in high school and that was years ago. It does irk me though that howstuffworks.com is giving out incorrect information, to say the least. I am still somewhat confused on this subject though, since what they say seems to make sense from a mathematical stand point but certainly not from a common sense stand point. If kinetic energy increases exponentially due to velocity, it would seem that a machine with a constant power of say 1000 watts would accelerate a car, say, at a slower rate at higher speeds since the kinetic energy required at higher velocities increases exponentially. What am I missing here? Is the answer that power would also increase at higher speeds, since more energy is being transferred from the view point of someone standing still, watching the car from the start of acceleration?

 

[Dale]

I read the following on howstuffworks.com, and their explanation just doesn't seem correct to me.
...
The way they explain this implies that it is more difficult for an object that is already in motion to accelerate further. Wouldn't this imply that a greater force is necessary to accelerate an object at higher speeds than at lower speeds? As far as I can remember F=ma and neither acceleration nor force is affected by current velocity (excluding velocities that are near the speed of light, of course).

The explanation is correct. It does in fact take more energy to go from 40 to 60 mph than from 0 to 20 mph. Remember, work is F.d. Let's say that you have a hypothetical automobile that can produce the same force at all speeds. By f=ma this vehicle will therefore take the same amount of time to go from 0 to 20 as it does to go from 40 to 60. During that same amount of time the vehicle will go a much greater distance from 40 to 60 than it did from 0 to 20. Even though f is the same the work will be much greater in the 40 to 60 case because d is much greater and W=f.d.

Also, doesn't kinetic energy depend on frame of reference as well? If you are already moving at 20 mph as in the example at a constant speed, couldn't your frame of reference just as easily be 0 mph? I realize that with the way a car's engine works and the way it is geared, air resistance, friction, etc, this may not be true, but what about if it was a spaceship in a perfect vacuum? It doesn't seem that if you had two identical rockets used for thrust that the first rocket used would accelerate you more or more quickly than the second rocket used does just because your velocity relative to Earth was now higher after firing off the first rocket.

Yes, kinetic energy is relative to the frame of reference. You can do this kind of analysis in any reference frame if you are careful enough, but in the car's case you have to take into account the KE of the Earth and in the rocket case you have to take into account the KE of the exhaust.