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?


1 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.

  • $\begingroup$ But am I right then? Because you reach 4.2 when you've stopped moving, aka the friction is 0, in other words, this is incorrect? $\endgroup$
    – user14445
    Commented Nov 5, 2012 at 17:29
  • 1
    $\begingroup$ 4.2 meters is the distance traveled while under the influence of the constant kinetic friction force. The work that this constant force does is correctly calculated. What happens at 4.2 meters is of no direct concern. You could calculate the work done by the time the car travels 3.5 meters or 2.1 meters, and the math would be valid for those distances traveled. What happens when the car reaches a given distance is not relevant, as long as the force was constant the whole time before it arrived there. $\endgroup$
    – Muphrid
    Commented Nov 5, 2012 at 17:32

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