Understanding Work and the conservation of energy

Question

We have a car with a mass of $780 kg$ with travels with a speed of $50 km/h$. The car brakes and after $4,2m$ is stops completely. Warmth is created. Calculate the friction.

I solved this easily, by simply filling in the data like a headless chicken (the solution I'm showing is from the correction model, I got the same answers but I did it without writing anything down, so):

$E_{total 1} = E_{total 2} $

$0.5mv_1^2 = 0.5mv_2^2 + Q = 0+ F_w . s $

$0.5 . 780 . (50/3.6)^2 = F_w . 4.2 $

$F_w = 1,8.10^4 N$

I didn't have any trouble with this, I got the same answer, but than I started thinking about it, and I am increasingly finding the solution illogical. The LHS is completely logical to me, but the right hand side. Normally, the formula of work is the resultant force times the distance d. But, after you've travelled the 4,2 meters, your $F_w$ is $0$. So how can you say that $F_w \times d$ = LHS, because by the time the total $d$ (4.2) is reached, the resistance has already turned into 0. What is the logic behind this? I know this is high school level so it is simplified, but even then, knowing it's simplified a lot, I don't understand the logic. Can someone explain?

Answer 1

Centered dots for multiplication in LaTeX are put in with \cdot

The force of kinetic friction does not gradually draw down to zero. It has a specific value for any nonzero velocity, and only at zero velocity does the friction force exhibit discontinuity as we abruptly enter the regime of static friction. Discontinuities in general are hard to fathom, but this is the standard way of teaching friction.

Think of it this way: the force on the car while it is moving is constant, so the work is easy to calculate. Once the car stops moving, the force may be different...but the car no longer moves, and so there is no work to account for.