Newton's force equation versus energy

I have a two-part question involving the difference between force and energy. Say you have two vehicles and one is 1000 kg and the other is 2000 kg. Both of these vehicles have the exact same electric motor and battery propelling them. Also, they have gearing such that both electric motors wind up the same as far as angular speed when they accelerate. And, road friction and aerodynamic drag are negligible. The 1000 kg vehicle would have twice the acceleration, correct? The second part of my question involves energy. I found that people on the Exchange believe that force has to somehow increase with velocity, because of increased energy. This example, at least in my mind, proves otherwise. In my example you're using the same force / electricity / energy to propel the two vehicles, but the 1000 kg vehicle is going to have twice the energy of the 2000 kg vehicle, at any point in their accelerations. Am I correct?

• The kinetic energy of the vehicle has to be equal to the work done by the motor less the work lost to aerodynamic drag, and the kinetic energy is proportional to the mass: $KE = \tfrac12 mv^2$. So to get the 2000kg vehicle to velocity $v$ requires at least twice the work needed to get the 1000kg vehicle to the same velocity $v$. Commented Dec 11, 2019 at 15:05
• I don't see how all these things can be true simultaneously. You can't have the same force and energy applied to the two different vehicles, and still have them have different energies at the end; if they start with the same energy.
– JMac
Commented Dec 11, 2019 at 17:03

A constant force $$F$$ on a mass $$m$$ implies the acceleration $$\ddot{x}=\frac{F}{m}$$, so you are right in your first answer. The velocity becomes $$\dot{x}= \frac{Ft}{m}$$, so the kinetic $$E_\mathrm{kin}$$ energy is

$$E_\mathrm{kin} = \frac12 m \Big(\frac{Ft}{m}\Big)^2=\frac12 \frac{F^2t^2}{m}$$.

Hence, you are also right in your second answer: the 1000 kg vehicle is going to have twice the energy as the 2000 kg vehicle.

• Comments are not for extended discussion; this conversation has been moved to chat. Commented Dec 12, 2019 at 18:34

You might be confusing two things:

As stated in answers above, the acceleration is going to be $$a=F/M$$ which means that the speed after a time $$t$$ is going to be $$v=Ft/M$$ and therefore the kinetic energy $$K={1\over 2}Mv^2 = {1\over 2} {F^2t^2 \over M}$$ So, after the same time the bigger vehicle will have twice the amount of kinetic energy.

However, this does not mean that the two "engines" have produced the same amount of energy! To accelerate the bigger vehicle the engine has consumed more energy (or more fuel, if you prefer). Indeed, in order to keep the force constant despite the difference in weight the engine needs to deliver more energy to the bigger vehicle. That engine has "worked" more!

• no, I don't think so. You and I have probably seen a dozen train examples where the engine, having a given power, accelerates at half the speed when you add enough train cars to double its mass. This example is no different. Commented Dec 16, 2019 at 15:02

The 1000 kg vehicle would have twice the acceleration, correct?

Correct. Per Newton's second law:

$$F=ma$$

$$a=\frac{F}{m}$$

Where $$F$$ is the net force acting on an object of mass $$m$$ and $$a$$ is the acceleration of $$m$$. So if both vehicles are subjected to the same net horizontal (to the road) force $$F$$, the 1000 kg vehicle will have twice the acceleration as the 2000 kg vehicle because it is twice the mass.

I found that people on The Exchange believe that Force has to somehow increase with velocity, because of increased energy.

Force does not increase with velocity. For a constant net force momentum increases linearly with velocity and kinetic energy increases with as the square of the velocity.

This example, at least in my mind, proves otherwise. In my example you're using the same Force / electricity / energy to propel the two vehicles, but the 1000 kg vehicle is going to have twice the energy of the 2000 kg vehicle, at any point in their accelerations. Correct?

It is correct that at any given time the smaller vehicle will have twice the kinetic energy of the larger. But it is because the same force is applied to each, not that the same electricity/energy is consumed by both. For conservation of energy the electricity/energy used by each vehicle for a given time has to equal their kinetic energies.

The following is one way to show how the kinetic energy of the smaller vehicle will be twice the larger:

For a constant acceleration, the velocity of each vehicle as a function of time will be $$v=at$$. Since the 1000 kg vehicle has twice the acceleration of the 2000 kg vehicle, at any given time the smaller vehicle will have twice the velocity of the larger vehicle. So let the velocity of the larger vehicle be $$v$$, then the smaller vehicle will be $$2v$$ at any given time, and the kinetic energy of each vehicle, where $$m$$ is the smaller vehicle and $$M$$ is the larger vehicle

$$KE_{M}=\frac{Mv^2}{2}$$

$$KE_{m}=\frac{(m)(2v)^2}{2}=\frac{4mv^2}{2}=2mv^2$$

Since $$m$$=$$\frac{M}{2}$$

$$KE_{m}=Mv^2$$

Therefore the kinetic energy of the smaller vehicle will be twice the larger vehicle.

UPDATE:

I don't understand how energy consumption of the lighter vehicle would be any greater than the larger if both electric motors are spinning at the same speed and applying the same Force to the drivetrain.

You are missing an important point. Although the kinetic energy of the smaller vehicle is twice the larger vehicle for a given elapse time $$t$$, the smaller vehicle also travels twice the distance than the larger for the same elapsed time.

The distance traveled by each vehicle in time $$t$$ is

$$d=\frac{at^2}{2}$$

Since the acceleration $$a$$ of the smaller vehicle is twice that of the larger, the smaller vehicle travels twice the distance for the same elapsed time.

The work $$W$$ done by the engine on each vehicle equals the force $$F$$ it applies times the distance $$d$$ traveled or

$$W=Fd$$

Therefore, since the force $$F$$ applied by the engine in each vehicle is the same, the work done on the smaller vehicle is twice that of the larger for the same elapsed time, or

$$W_{m}=2W_{M}$$

Which is the reason the smaller vehicle has twice the kinetic energy of the larger vehicle for the same elapsed time.

An underlying principle here is the work-energy theorem which states that the net work done on an object equals its change in kinetic energy.

Hope this helps.

• You're saying its correct that you're using "the same force/energy" and that "but the 1000 kg vehicle is going to have twice the energy of the 2000 kg vehicle". This seems to violate energy conservation... the two vehicles don't use the same energy.
– JMac
Commented Dec 11, 2019 at 17:07
• @JMac No I did not intend to say "the same force/energy" is correct. When I said correct I was referring to the fact that after the same amount of time, the smaller vehicle will have twice the KE as the larger. I should have separately addressed the former. Will point out. Thanks. Commented Dec 11, 2019 at 17:14
• It's typically something I wouldn't notice, I just get the feeling that was a crucial part of what the OP is actually trying to get at. I knew you weren't suggesting otherwise, just better to clear it up directly.
– JMac
Commented Dec 11, 2019 at 17:16
• @JMac Fixed. Thanks again. Commented Dec 11, 2019 at 17:23
• @JamesMontagne Sorry James, but JMac is correct. Neglecting all dissipative effects (air friction, mechanical friction, etc., etc.,) the electrical energy used by each car has to equal its kinetic energy. No getting around energy conservation. The only thing that is the same is the force applied by each. But force does not equal energy. If you take another look at my revised answer (paragraph following the last yellow highlighted) you will see I pointed this out. Commented Dec 16, 2019 at 18:50