Minimum Coefficient of Kinetic Friction Question.

Question

idku · Answer

You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks that have lost their brakes on mountain grades.

Typically such a lane is $\rm horizontal$ (if possible) and
about $\color{red}{37.0~\rm m}$ long. We can think of the ground as
exerting a frictional drag force on the truck. The truck
enters the gravel lane with a speed of $\color{red}{24.59~\rm m/s}$.

Calculate the minimum coefficient of kinetic friction between the truck and the lane to be able to stop the truck.

idku · Answer

@IrishBoy123 if you have some time, please ...

pawanyadav · Answer

I think ....
Total K.E = work done by friction 
From above eq. you can find coefficient of kinetic friction.

irishboy123 · Answer

yes, i would agree!

$F \times x = \mu mg x = \frac{1}{2}m v^2$

so you don't worry about mass

idku · Answer

Oh, I am thinking it makes sense to me

idku · Answer

$\frac{1}{2}m\cdot v^2=\mu_k\cdot m\cdot g\cdot x$
$\frac{1}{2} 24.59 ^2=\mu_k\cdot  9.80\cdot 37$

idku · Answer

And that coefficient of friction is minimum, because it is the smallest (causing the truck to stop right at the end)

irishboy123 · Answer

yeah, cool!

plus, just give it a bit of this: $\mu = \dfrac{{1 \over 2} v^2}{gx}$

from the equations we did.  it'll make your life easier

idku · Answer

My problem is that I don't (didn't) know that kinetic energy = work done by friction,
and I didn't recall that $f_k=\mu_kN=\mu_kmg $

idku · Answer

tnx for your response.

I got the correct (3 Sig Figs) answer;)
μ=0.834