Thermal energy dissipated from the brakes in a car

[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?

 

[simphys]

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

 

[malawi_glenn]

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

You did you define ΔK and ΔU?

K_initial+U_initial > K_final+U_final
Thus
K_initial+U_initial = K_final+U_final + W
where W is the work done by friction (positive here)

 

[simphys]

You did you define ΔK and ΔU?

K_initial+U_initial > K_final+U_final
Thus
K_initial+U_initial = K_final+U_final + 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 $\Delta K = W$ --> $\Delta K = W_C + W{NC}$ --> $\Delta K = -\Delta U + W_{NC}$ --> $\Delta K + \Delta U = W_{NC}$

 

[malawi_glenn]

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

 

[simphys]

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

 

[malawi_glenn]

okay, apologies. change in kinetic energy aka the work-energy principle

I meant
is ΔK = K_initial - K_final or ΔK = K_final - K_initial
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.

 

[simphys]

I meant
is ΔK = K_initial - K_final or ΔK = K_final - K_initial

always $\Delta K$= K_final - K_initial
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 $2$.

 

[simphys]

I meant
is ΔK = K_initial - K_final or ΔK = K_final - K_initial
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.

didn't see your explanation aferwards.
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.

 

[kuruman]

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.

 

[simphys]

The statement of the problem is missing the initial and final velocities. Please provide them.

apologies edited

 

[simphys]

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

 

[malawi_glenn]

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

@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.

 

[simphys]

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)

 

[malawi_glenn]

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]

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.

 

[malawi_glenn]

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]

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.

 

[simphys]

Now you have be meticulous again, what kind of energy are you referring to?

you're right.. potential energies
Apologies.

 

[malawi_glenn]

you're right.. potential energies

So if kinetic energy is changed, no work is done?

 

[simphys]

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?

 

[simphys]

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 $\Delta U =$ U_final - U_initial
to saying that $W_{NC} = -\Delta E_{TH}$ Or what other definition would that be?

 

[malawi_glenn]

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

 

[simphys]

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

 

[malawi_glenn]

change in energy

what energy?
Have you not covered the concept of mechanical energy in your book?

 

[simphys]

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

 

[kuruman]

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]

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.