1
$\begingroup$

For a long time, I have been thinking about this: In order to stop a moving body, I have to apply impulse to change the existing momentum to $0$. For the impulse I have to impart force on the moving body . That means I have to eliminate all the kinetic energy associated with the moving object in order to bring it to a halt. Thus, work is done on the object, right? If work is done, energy must be given by me...

Here is where my confusion arises:

  1. If I eliminate the associated kinetic energy of the body, where does this energy go?
  2. The work(?) done by me on the moving object to stop it obviously uses energy from me. Again, where does this energy go?
$\endgroup$
4
$\begingroup$

Thus, work is done on the car, right?

No, the car does (positive) work on whatever is stopping it.

Alternatively, you could say that negative work is done on the car, but still, the meaning is the same: the car loses energy and something else gains that energy. What that something else is, and what type of energy it gains, depends entirely on how the car is stopped.

$\endgroup$
2
$\begingroup$

If there was no friction, your breaks couldn't clamp down on your rotors to slow the car, and the car's tires couldn't "stick" to the pavement. Your engine is generating energy, and it does cause pressure in the braking system. That braking system, triggered by you and your foot (and assisted by the car) converts that motion to heat through friction between your break-pads and rotors. That's where the energy to stop comes from, it's actually just a drastic increase in friction against the forward motion of the car.

$\endgroup$
2
$\begingroup$
  1. If I eliminate the associated Kinetic Energy of the car, where does this energy go?

While conservation of energy dictates that the energy due to the car's motion must be conserved, it does not say how. An old-style braking system converts that kinetic energy into heat. A more modern regenerative braking system converts that kinetic energy into a form that can be used later.

  1. The work(?) done by me on the car to stop it obviously uses energy from me. Again, where does this energy go?

Hardly any of the work comes from you, the driver of the vehicle. In the case of a non-regenerative braking system, the work (which is negative) comes from the Earth. In the case of a regenerative braking system, the car's kinetic energy is used to drive the car's electric motor in reverse, as if it was an electric generator. The generated electricity in turn is used to recharge the car's battery.

$\endgroup$
1
$\begingroup$

Unless you are stopping a moving body by some system specifically designed to convert the kinetic energy into useful potential energy, as in regenerative braking, in most cases the original linear kinetic energy of the object's center of mass gets converted to heat which raises the temperature of the surroundings by some small amount. This would be true in stopping due to friction, for example, as well as inelastic collisions where the two colliding bodies both come to rest in the frame of their mutual center of mass (in a collision some energy will initially be converted to sound waves rather than heat but the energy carried by the sound waves will itself disperse into heat as the sound dies out, likewise if the collision generates photons those photons will mostly be absorbed by atoms in the surroundings and converted into heat, assuming the collision takes place on Earth).

The main thing to understand here is that in statistical mechanics, the molecules that make up substances are constantly moving around in different directions, and colliding with each other and transferring energy between them, with the momentum of different molecules pointing in many different directions at any given moment, so that the vector sum of all the little momentum vectors can be zero in the substance's center-of-mass frame. So in a collision, the initial linear kinetic energy of the colliding object's center-of-mass can be just go to increasing the energy of these thermal motions of the individual molecules, which aren't visible on large scales. So, the basic answer is that in this type of case, linear kinetic energy of the objects' center of mass before the collision is converted into thermal energy.

Each molecule will have a number of "degrees of freedom" that can contain a certain amount of energy at each moment. For example, for a diatomic molecule made of two atoms, its velocity vector can be broken down into three components (the components along the x, y, and z axes of a spatial coordinate system), so that's three degrees of freedom which can contain kinetic energy. Then there are two axes the molecule can rotate around, giving two more degrees of freedom that can contain kinetic energy, and the molecule can also vibrate with the distance between the two atoms that make it up changing slightly, which gives an additional degree of freedom for the kinetic energy of the back-and-forth vibration, and finally one more degree of freedom for the changing potential energy in the bond between the atoms (similar to two masses with a spring between them whose length is oscillating). An important result in statistical mechanics is the equipartition theorem, which says that on average, each degree of freedom will contain an energy of $(1/2)kT$, where $T$ is the temperature of the system and $k$ is Boltzmann's constant. (However, some degrees of freedom may not "become available" until a certain temperature threshold is passed--for example, for air at room temperature, very little energy is stored in vibrational degrees of freedom because the molecular collisions aren't powerful enough to cause much vibration, so diatomic air molecules at room temperature effectively have five degrees of freedom rather than seven.)

So if you have a substance composed of $N$ identical molecules which each have $f$ degrees of freedom, the total thermal energy $U$ in the substance should be given by $U = (f/2)NkT$. In an inelastic collision on Earth, while there will initially be a localized increase in the thermal energy of the colliding objects and their immediate surroundings, this will soon be dispersed into a much larger number of molecules in the surrounding air and ground. But if you imagine two objects colliding inelastically in space, with a negligible amount of energy lost to photons, then you can actually use this to predict the increase in temperature of the objects after the collision, since the increase in thermal energy must be equal to the decrease in linear kinetic energy of the two objects in their mutual center-of-mass frame.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy