Calculating Work Done by Brakes: A Step-by-Step Guide

[Chica1975]

Homework Statement

the driver of a car mass 1368kg braked momentarily to slow down from 92km/h (25.556m/s) to 63 km/h (17.5m/s) how much work was done by the brakes (friction).

Homework Equations

w = Fd
F=ma
difference between initial acceleration and final acceleration s 8.056m/s (25.556 - 17.5)
mg = 13406.40

The Attempt at a Solution

I ave drawn the picture, I have tried a number of calculations - I am completely lost - I have no friction coefficient and don't know where to start. I have looked a book I have that has problems but has nothing like this. Apparently, the answer is 237.

 

[willem2]

Try to use conservation of energy. I have no idea where 237 comes from, it's much to low (and has no units!)

 

[Chica1975]

Thanks - this is the answer the lecturer has given for this question. I have no idea how to get it out.

 

[Chica1975]

the units are KJ

 

[Chica1975]

Thanks I tried to use conservation energy equation. I can't get this out.

Please can some explain to me step by step how to do this - I have been on this question for the last hour and a half and have tried all sorts of things with different equations - its not working.

I need a reference point to start from I have no idea - I am stuck.