# Energy loss without work or heat?

Suppose we have a mass-spring system in equilibrium as in the above part of the picture(it's a frictionless surface).

Then I attached a weightless rope to the mass from one end and a car from the another end. This car contains fuel(chemical energy). I turn on the car and drive it to the right, so the motor of the car is going to apply force to the right and displace the mass, with the Work done on the mass being equal to: $U=1/2kx^2$.

Now if I cut the rope, the potential energy stored in the mass is going to be converted into a kinetic one, and the mass is going to undergo a simple harmonic motion.

But suppose I want to maintain this present configuration(the part of the picture below), that is, I want to keep the mass at rest away from the equilibrium position, then I will have to keep pressing on the pedal of the car so as to apply a force that continuously counteract and cancel out with the force of the spring. As a result the fuel is continuously being burnt and chemical energy is lost, Although the mass is at rest and no Work is done at all.

My confusion is this: Although there is no Work done at all, this system continuously losses energy.

what is wrong here?

I'm well aware of the fact that there are other energy transfer mechanisms like heat, but assume that the engine of the car is an ideal engine.

Then how this system losses its energy?

• An ideal engine would not be burning fuel when the car isn't moving. (Also, how is it supposed to move the car in the first place on the frictionless surface?)
– user21968
Jun 16, 2015 at 19:24
• Wheels will not work on a frictionless surface... Jun 16, 2015 at 19:58

(Wheels will not work on a frictionless surface... But let's assume you mean that surface under the box is frictionless and not the surface under the car)

then I will have to keep pressing on the pedal of the car so as to apply a force that continuously counteract

Something is wrong here. For equilibrium and a non-moving box, you don't want the box or car to move. So, why press the pedal? Why not just apply the handbrake? The car will stand still from static friction.

By pushing the pedal down you risk wheelspin - then kinetic friction takes over, which is always less than the static friction, and the force is then even smaller.

If it will not stand still at brakes applied, then the friction on the surface is not enough to hold the equilibrium.

If is does stand still even when pushing the pedal, then the engine is not applying enough work to make the wheels turn (it must overcome the static friction). Any fuel burnt is turned into heat within the car.

Bottom line is: No energy is lost to keep the equilibrium if the friction is big enough. If the engine is running, then the burnt fuel is converted into heat within the car.

If you "push the pedal" (presumably the accelerator) and the car isn't moving, then 2 things are happening.

1) The engine is burning fuel, the pistons are going up and down, and the crankshaft is turning. All of this takes energy.

2) But the wheels aren't moving. This says that something is interrupting the transfer of motion from the crankshaft to the wheels. That something is a clutch, and it's slipping, and friction is causing it to heat up. That heat energy is not being delivered to the wheels, either.

The sum of the two energies, plus engine inefficiency (hot exhaust gas, piston and bearing friction, etc) all add up to the energy being released by the burning fuel.

So the kicker in your question is "assume that the engine of the car is an ideal engine". If you do that, no fuel is burned and no energy wasted. Either that, or all the energy is wasted in the clutch.

Another approach to the "perfect engine" is a superconducting electric motor. As you push the pedal, you increase the current in the motor, but the engine does not turn. Under these circumstances, there is no energy lost in the motor, since (it being a superconductor and there is no change in the motor's magnetic fields) the voltage across the motor is zero, and the power being dissipated in the motor (voltage times current) is also zero.