How Much Energy is Dissipated When Braking from 75 to 55 mph?

[thetexan]

Here's the scenario...

I'm driving down the highway at 75 mph and not paying attention. I suddenly see a 55 mph speed limit sign and hit the brakes to slow down. I decelerate from 75 to 55 in 3 seconds. The car weighs 4700 lbs and has 4 disc brakes.

Now assuming the only factor is the friction of the brakes (no wind resistance, etc.) do I have enough info to calculate how much energy was dissipated by the 4 brakes and determine if that was all in heat and if so how much heat? If not precisely, a good estimate?

tex

 

[PWiz]

If the deceleration was constant, then yes, it can easily be calculated. Just apply conservation of energy.

 

[gleem]

what if the deceleration was not constant?

 

[nasu]

Still can do the same thing.
If the only force acting was friction, all the kinetic energy lost is dissipated by friction.
All you need is the initial and final KE.

 

[rootone]

what if the deceleration was not constant?

I guess using the average deceleration while applying the brake would get you the same or a very close to same result.

 

[nasu]

You don't need any deceleration to answer the question in the OP.

 

[PWiz]

what if the deceleration was not constant?

You would need a function $v(t)$ which describes the velocity of the car with time. The force can easily be represented by $F=m\frac{dv}{dt}$. Kinetic energy is given by $E=\int_{s_1}^{s_2} Fds$. Can you guess how to represent F in terms of s? If yes, then you'll see that the solution is just as straight forward .

 

[Khashishi]

You don't need that. Just take initial energy - final energy.

 

[PWiz]

You don't need that. Just take initial energy - final energy.

But the OP asked for a calculus approach.

 

[Drakkith]

But the OP asked for a calculus approach.

Where? I see nothing about that in the original post.

 

[PWiz]

Where? I see nothing about that in the original post.

Whoops, I guess I mixed that question up with some other one
@OP If you don't have to show steps of calculus, then you can simply use the formula $\frac{1}{2}mv^2$, and you have all the information you need. (Actually this formula is derived from calculus since $\int_0^{s} Fds= m \int_0^s \frac{dv}{dt} ds= m \int_0^v v dv$, which yields the Newtonian expression above) Just remember that the energy will always be converted into some other form.