Kinetic Energy of Cars: Does 4x Energy Needed for 0-20kph?

[deanbo]

For example if a car accelerates from 0 to 10 kph it creates 50,000 joules of kinetic energy? If a car acclerates from 0 to 20 kph it creates 200,000 joules of energy. Does this mean that four times the amount of energy is required from the cars engine to go from 0 to 20 kph instead of 0 to 10 kph? Or is this kinetic energy the result of some other type of conversion?

 

[Nabeshin]

Not exponentially, just quadratically. And the energy obviously comes from however you doubled the car's velocity (people pushing, the engine, an explosion, etc.).

 

[russ_watters]

And you'll notice this manifesting in a car accelerating from 30-60 much slower than from 0-30.

 

[deanbo]

Not exponentially, just quadratically. And the energy obviously comes from however you doubled the car's velocity (people pushing, the engine, an explosion, etc.).

Thanks! What I am also trying to determine is all the kinetic energy that is generated by the cars movement simply a result of the acceleration (velocity) itself. I'm guessing not otherwise mass would not be part of the formula for working out the amount of kinetic energy produced?

 

[mikelepore]

The formula for KE comes from assuming that conservation of energy will be obeyed, so the work done to accelerate the car will be equal to its change in KE. Begin with this equation from kinematics: vf^2 - vi^2 = 2ax, where vi is initial velocity, vf is final velocity, a is acceleration, x is displacement. Multiply each term by the object's mass m. Replace ma by force F. Change the 2 on one side to a (1/2) on the other side. Replace Fx by work W. Now, if you define the term (1/2)mv^2 to be something called KE, your equation will say this: work done = final KE - initial KE. This was just derived only for the case where the force was constant over the displacement, making it legal to say work W = Fx instead of using calculus, but since conservation of energy is a general law, the expression for KE is general also.