# Thermal energy dissipated from the brakes in a car

• simphys
In summary: It only becomes negative once ##W_{NC}## is changed with ##F*displacement## hence that is why I use it like...In summary, the kinetic energy of the car decreases due to the braking process.
simphys
Homework Statement
96. Proper design of automobile braking systems must account
for heat buildup under heavy braking. Calculate the thermal energy dissipated from brakes in a 1500-kg car that descends a 17° hill. The car begins braking when its speed is ##95km/h## (##=26.39m/s##)and slows to a speed of ##35km/h##(##=9.72m/s##) in a distance of 0.30 km measured along the road.
Relevant Equations
conservation of enertgy
My sign doesn't check out and I don't get why that'd be the case.
Forces that act --> ##F_{fr} and F_g##
derivation:
##\Delta K = W_{NC} + W_C (1)##
##\Delta K + \Delta U = W_{NC}##
##\frac 12 mv_2^2 - \frac 12 mv_1^2 + mgy_2 - mgy_1 = W_{NC}## NOTE: ##y_2## assumed to be datum line so ##y_2 == 0##
filling in the data I get:
##W_{NC} = -1.742E6 = E_{th}## instead of ##1.742E6 ##

Do I need to make ##W_{NC}## to ##-W_{NC}## in ##(1)## perhaps or?
or..
can I just say ##W_{NC} = -1.742E6## which is negative work done by the friction force.
From this we are able to conclude that the magnitude of the Thermal energy dissipated ##E_th = 1.742E6##. Is this permitted or do I actually need to get a + sign in the equation itself?

Last edited:
I would assume that this is correct since it indicates that the friction force does negative work ?

simphys said:
I would assume that this is correct since it indicates that the friction force does negative work ?
You did you define ΔK and ΔU?

Kinitial+Uinitial > Kfinal+Ufinal
Thus
Kinitial+Uinitial = Kfinal+Ufinal + W
where W is the work done by friction (positive here)

simphys
drmalawi said:
You did you define ΔK and ΔU?

Kinitial+Uinitial > Kfinal+Ufinal
Thus
Kinitial+Uinitial = Kfinal+Ufinal + W
where W is the work done by friction (positive here)
well I said that ##W_C = -\Delta U## but didn't write it.
And yeah I get that it is like that, but there was one exercise where the work done by the non concservative forces came on the side of the initial energy which confused me (a couple days ago) and since then I use ##\Delta K = W## --> ##\Delta K = W_C + W{NC}## --> ## \Delta K = -\Delta U + W_{NC}## --> ##\Delta K + \Delta U = W_{NC}##

I always write down explicitly what i mean by ΔK etc when I solve problems. I think its a good habit.

simphys
drmalawi said:
I always write down explicitly what i mean by ΔK etc when I solve problems. I think its a good habit.
okay, apologies. change in kinetic energy aka the work-energy principle

simphys said:
okay, apologies. change in kinetic energy aka the work-energy principle
I meant
is ΔK = Kinitial - Kfinal or ΔK = Kfinal - Kinitial
I have marked soo many physics students exams where they mess up on this particular matter, since they have not written down what they mean by e.g. ΔK and then suddenly midway through their solutions seems to switch definition in their heads and get wrong result.

drmalawi said:
I meant
is ΔK = Kinitial - Kfinal or ΔK = Kfinal - Kinitial
always ##\Delta K ##= Kfinal - Kinitial
might not be so clear with the subscripts my apologies. I normally have my picture accompanied with the problem at which I show position ##1## and ##2##.

drmalawi said:
I meant
is ΔK = Kinitial - Kfinal or ΔK = Kfinal - Kinitial
I have marked soo many physics students exams where they mess up on this particular matter, since they have not written down what they mean by e.g. ΔK and then suddenly midway through their solutions seems to switch definition in their heads and get wrong result.
Well no I know that it is the second option, but what I didn't get is, why is the way I wrote it down not correct
I mean to me it kind of makes sense that ##W_{NC}## is negative because it's friction aka opposite to the direction of motion.

simphys said:
Homework Statement:: 96. Proper design of automobile braking systems must account
for heat buildup under heavy braking. Calculate the thermal energy dissipated from brakes in a 1500-kg car that descends a 17° hill. The car begins braking when its speed is and slows to a speed of in a distance of 0.30 km measured along the road.
The statement of the problem is missing the initial and final velocities. Please provide them.

malawi_glenn
kuruman said:
The statement of the problem is missing the initial and final velocities. Please provide them.
apologies edited

@drmalawi let me show you a small piece of the textbook to why I do it this way

malawi_glenn
simphys said:
I mean to me it kind of makes sense that WNC is negative because it's friction aka opposite to the direction of motion.
if indeed ΔK = Kfinal - Kinitial and ΔU = Ufinal - Uinitial then your WNS will be negative, and the absolute value of this is the amount of thermal energy
simphys said:
@drmalawi let me show you a small piece of the textbook to why I do it this way
You don't have to I know what you mean now, I just wanted to give you a little advice on being meticulous regarding definitions

simphys
if this is not clear, I can make a picture from my actualy textbook if necessary, this is an online pdf version

as can be read here, It only becomes negative once ##W_{NC}## is changed with ##F*displacement## hence that is why I use it like this.

drmalawi said:
if indeed ΔK = Kfinal - Kinitial and ΔU = Ufinal - Uinitial then your WNS will be negative, and the absolute value of this is the amount of thermal energy

You don't have to I know what you mean now, I just wanted to give you a little advice on being meticulous regarding definitions
so if I assume it as the absolute value, that'll be a correct way to solve the problem? (just to make sure)

simphys said:
assume it as the absolute value
you don't have to assume it, it is by definition. The work done by friction will reduce the mechanical energy in your system. The lost mechanical energy is converted to heat.

simphys
drmalawi said:
if indeed ΔK = Kfinal - Kinitial and ΔU = Ufinal - Uinitial then your WNS will be negative, and the absolute value of this is the amount of thermal energy

You don't have to I know what you mean now, I just wanted to give you a little advice on being meticulous regarding definitions
Didn't see the second comment again.. 'You don't have to I know what you mean now, I just wanted to give you a little advice on being meticulous regarding definitions' Thank you. for this specific chapter about conservation of energy, the most important thing that I overlooked was actually the biggest relation that is made. which is the fact that during a change in energy there's always work done in the process. I looked at it at first (work and energies) as being two different ways to solve a problems whilst they're very closely related. It's just that we convert the work done by forces into energies dependent on what system we would choose.

simphys said:
a change in energy there's always work done in the process.
Now you have be meticulous again, what kind of energy are you referring to?

simphys
drmalawi said:
you don't have to assume it, it is by definition. The work done by friction will reduce the mechanical energy in your system. The lost mechanical energy is converted to heat.
oh okay great, thanks a lot for that!
Yep that I know, but you know that in physics a - or + signs make a big difference and hence why I wanted to make sure that it'd be correct to do so after getting a negative work done.

drmalawi said:
Now you have be meticulous again, what kind of energy are you referring to?
you're right.. potential energies
Apologies.

simphys said:
you're right.. potential energies
So if kinetic energy is changed, no work is done?

simphys
drmalawi said:
So if kinetic energy is changed, no work is done?
okay let me rephrase:
No, now that I think about it.. this is valid for all of the energies.
Whenever there's work done in a process, there is energy transformed either from one form to another or from one object to another.
Conclusion: to say simply 'energy' suffices then, or am I wrong?

drmalawi said:
you don't have to assume it, it is by definition. The work done by friction will reduce the mechanical energy in your system. The lost mechanical energy is converted to heat.
additional: sorry to bring this up again.. but I just thought about it.
What do you mean by definition? Is it similar to saying that for ##-\Delta U = W_C## where ##\Delta U = ## Ufinal - Uinitial
to saying that ##W_{NC} = -\Delta E_{TH}## Or what other definition would that be?

If you drop a ball from an height (ignore air resistance). Work is done on the ball, but its mechanical energy is constant.

simphys
drmalawi said:
If you drop a ball from an height (ignore air resistance). Work is done on the ball, but its mechanical energy is constant.
So there is only change in energy when a non-conservative force is acting
nooo nvm.. because there is a change in kinetic energy and potential

simphys said:
change in energy
what energy?
Have you not covered the concept of mechanical energy in your book?

simphys
drmalawi said:
what energy?
Have you not covered the concept of mechanical energy in your book?
yeah no no, what I am talking about (/having in mind) are the energies that are associated with the work done by forces and additionally kinetic energy aka the energy of motion.
Not ME energy which is composed of 2 energies i.e. potential and kinetic

A lot of ambiguity is avoided if one thinks of the Earth-roller coaster as an isolated system, i.e. no external forces do work on any of its components, which is the case here. Then total, not just mechanical, energy K+U, is conserved. The energy conservation equation is written in terms of differences, the sum of which is zero. In this case we have changes in kinetic, potential and thermal energy of the system $$\Delta K+\Delta U+\Delta E_{\text{therm}}=0$$ The changes are, of course, always ##Final - Initial##. When one puts the numbers in., the change in thermal energy comes out positive because the temperature of the brakes, which are part of the system, rises without a corresponding decrease of temperature elsewhere in the system.

I note that this view is consistent with the statement in the textbook excerpt, end of next to last paragraph: "Total energy is conserved." In my opinion, total energy conservation is much easier to implement than having to deal with work done by conservative and non-conservative forces.

simphys
kuruman said:
A lot of ambiguity is avoided if one thinks of the Earth-roller coaster as an isolated system, i.e. no external forces do work on any of its components, which is the case here. Then total, not just mechanical, energy K+U, is conserved. The energy conservation equation is written in terms of differences, the sum of which is zero. In this case we have changes in kinetic, potential and thermal energy of the system $$\Delta K+\Delta U+\Delta E_{\text{therm}}=0$$ The changes are, of course, always ##Final - Initial##. When one puts the numbers in., the change in thermal energy comes out positive because the temperature of the brakes, which are part of the system, rises without a corresponding decrease of temperature elsewhere in the system.

I note that this view is consistent with the statement in the textbook excerpt, end of next to last paragraph: "Total energy is conserved." In my opinion, total energy conservation is much easier to implement than having to deal with work done by conservative and non-conservative forces.
Exactly.. that is what got me confused the first time round. I mean I get it now, but yeah.. You see energies and all of a sudden a work done, not considered as an energy, that was the confusing part back then. It's dependent on the system that you have chosen, but hey try to figure that out when it hasn't been stated explicitly, or I would say stated once kind of vaguely and not repeated.

And that's why I actually reverted back to the work-energy principle, I start from there and see which forces can be converted into an energy. i.e. the conservative forces.

Next time if asked such a question I'll just consider the system that also exerts that not-conservative force to find out what the ##\Delta E_{TH}## is. So thanks for that.

kuruman

## 1. How does thermal energy dissipate from the brakes in a car?

When a car is in motion, the brakes are applied to slow down or stop the car. This creates friction between the brake pads and the rotors, converting kinetic energy into thermal energy. The heat generated from this friction is then dissipated into the surrounding air, cooling down the brakes and allowing them to function properly.

## 2. What factors affect the amount of thermal energy dissipated from the brakes?

The amount of thermal energy dissipated from the brakes depends on several factors, such as the speed of the car, the weight of the car, the condition of the brake pads and rotors, and the type of braking system (e.g. disc brakes vs. drum brakes). Additionally, external factors like the temperature and humidity of the environment can also affect the dissipation of thermal energy.

## 3. Can thermal energy dissipation from the brakes cause damage to the car?

Excessive heat generated from the brakes can potentially cause damage to the car, such as warping or cracking of the rotors. This is why it is important to properly maintain and replace worn-out brake pads and rotors to prevent overheating and potential damage to the car.

## 4. Is thermal energy dissipation a form of energy loss in a car?

Yes, thermal energy dissipation from the brakes is a form of energy loss in a car. This is because the kinetic energy that was used to move the car is converted into thermal energy and dissipated into the environment, resulting in a decrease in the car's overall energy.

## 5. How can the dissipation of thermal energy from the brakes be reduced?

The dissipation of thermal energy from the brakes can be reduced by using high-quality brake pads and rotors, maintaining proper tire pressure, and avoiding excessive braking. Additionally, techniques such as coasting and downshifting can also help reduce the need for frequent braking and therefore decrease the amount of thermal energy dissipated from the brakes.

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