# How much energy is spent?

1. Mar 30, 2015

### 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

2. Mar 30, 2015

### PWiz

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

3. Mar 30, 2015

### gleem

what if the deceleration was not constant?

4. Mar 30, 2015

### 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.

5. Mar 30, 2015

### rootone

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

6. Mar 30, 2015

### nasu

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

7. Mar 30, 2015

### PWiz

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 .

8. Mar 30, 2015

### Khashishi

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

9. Mar 31, 2015

### PWiz

But the OP asked for a calculus approach.

10. Mar 31, 2015

### Staff: Mentor

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

11. Mar 31, 2015

### PWiz

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.