Does the work done by friction take into account the energy lost as heat in a system?

[manohman]

So friction as a nonconservative force, is path dependent when it comes to how much work is lost from a system right?

What confuses me however is understanding what that means, in terms of energy. So the work done by friction includes the energy that was neeeded to stop an obect (like a braking car) PLUS the heat released during the transition right?


Its weird because it means we assume that the friction force times the distance takes into account heat as well. I thought heat didnt count.

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For reference, this is the original question that cited mine.

A moving car suddenly activates its brakes. Initial conditions = before brakes activates. Final Conditions = at rest.

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[Odi]

The heat that results from breaking does not add any energy to the system. The heat that results from breaking is the actual transformation of kinetic energy into heat mediated by friction. The heat dissipates into the surroundings.

From the equation perspective of a car braking to a full stop:

1/2mv^2 = - mgUk*d

Nowhere is heat taken into account for any calculations.

 

[DrknoSDN]

So the work done by friction includes the energy that was neeeded to stop an obect (like a braking car) PLUS the heat released during the transition right?

There is no heat released here so there is a misunderstanding in the scenario. Friction does not always produce heat in an "ideal" scenario.

 

[Odi]

View attachment 187003

This question can be interpreted differently but a rotating and translational moving tire has both rotational kinetic energy and translational kinetic energy.

I'm assuming "just before the breaks" means the tire has no more rotational kinetic energy and purely translational kinetic energy. In this case I'd pick B.

If "just before the breaks" means the tire possessed some rotational kinetic energy for a fraction of time before the breaks were pressed, I'd pick A.

What's the answer?

 

[manohman]

This question can be interpreted differently but a rotating and translational moving tire has both rotational kinetic energy and translational kinetic energy.

I'm assuming "just before the breaks" means the tire has no more rotational kinetic energy and purely translational kinetic energy. In this case I'd pick B.

If "just before the breaks" means the tire possessed some rotational kinetic energy for a fraction of time before the breaks were pressed, I'd pick A.

What's the answer?

THe answer was B.

Oh i see so the Work done by friction takes into account both the energy lost by kinetic energy to stop the car (component of mechanical energy in the system that is conserved) AND the energy released as heat (if heat is released) which is the component of mechanical energy that is not conserved.

So if there was no energy released as heat, the answer would still be the work done by friction right? IM trying to understand then, would the car take longer to stop or a shorter period of time?

 

[Odi]

THe answer was B.

Oh i see so the Work done by friction takes into account both the energy lost by kinetic energy to stop the car (component of mechanical energy in the system that is conserved) AND the energy released as heat (if heat is released) which is the component of mechanical energy that is not conserved.

So if there was no energy released as heat, the answer would still be the work done by friction right? IM trying to understand then, would the car take longer to stop or a shorter period of time?

There is no energy lost as heat the way the question is asked.

Think of a rolling cylinder. It has rotational kinetic energy (from angular velocity and inertia) and translational kinetic energy (from linear velocity and mass).

"Just before applying the breaks" means once the breaks are stepped on there is no force being exerted by the car to produce a torque on its tires to cause a forward acceleration. It is a pure rolling motion with purely translational kinetic energy.

Now what force causes the car tire come to a stop?

A rolling tire has one single contact point with the ground. As the tire rolls, there is always a new contact point with the ground. In this case, you actually have static friction acting on every successive contact point with the ground as it rolls and NOT kinetic friction.

This static friction converts the initial translational kinetic energy which we start with into rotational kinetic energy; such that the static friction is producing a torque OPPOSITE the direction of rotation of the wheel, which causes an angular deceleration and ultimately a complete stop (see awesome paint job)

If the car tire skids, that's a whole other ball game. Imagine pulling a tire by the string... it would not roll, but skid across the road causing a kinetic friction now to slow it down. This is the case where kinetic friction would convert all the initial kinetic energy of the moving tire into heat. The exact same thing that happens when you apply an instantaneous force on a block and and send it in motion on a surface with friction. The question makes sure to say this is not the case.

 

[DrknoSDN]

@Odi has an excellent explanation but for this question you just need to read the information given and not extrapolate what would occur in reality. In reality their would be friction from the brake pad to the rotating wheel which causes heat (regardless of static or kinetic friction between the wheel and the ground), but that is not provided or mentioned so it can be ignored.

 

[manohman]

@Odi has an excellent explanation but for this question you just need to read the information given and not extrapolate what would occur in reality. In reality their would be friction from the brake pad to the rotating wheel which causes heat (regardless of static or kinetic friction between the wheel and the ground), but that is not provided or mentioned so it can be ignored.

I see. So if there's no energy lost as heat, then all the work done by friction is the where the kinetic energy goes when the brakes are applied, right?

So initially we have KEmax/initial. After the brakes are applied and the car comes to a stop, the kinetic energy is zero since it has been transformed, equal in magnitude to the work done by friction.

If we were to take heat into account, would we still say that the work done by friction is equal to the initial kineticenergy (both translational and rotational)? So the work done by friction always includes both the heat dispersed due to friction and the energy that went into stopping the object?

 

[Odi]

lol we're down the rabbit hole...

So initially we have KEmax/initial. After the brakes are applied and the car comes to a stop, the kinetic energy is zero since it has been transformed, equal in magnitude to the work done by friction.

This^^^^ is all you need to know

 

[DrknoSDN]

If we were to take heat into account, would we still say that the work done by friction is equal to the initial kineticenergy (both translational and rotational)? So the work done by friction always includes both the heat dispersed due to friction and the energy that went into stopping the object?

Yes, if you were to look at the total kinetic energy of the object (not just translational) then friction must account for the change in KE and the change in T.

Still, Odi only mentioned rotational energy to diagram where the friction force was being applied. It's not relevant to the question. If rotational energy was tested it would only be to ascertain if you know that when mass is shifted to the perimeter the object has more rotational inertia.

 

[manohman]

haha Awesome thanks guys. Really cleared up a big ambiguity for me and ah it feels a lot better when it makes sense. Cheers