Answer
$d=12m$
Work Step by Step
We know that
$\frac{1}{2}mv_1^2+\Sigma U_{1\rightarrow 2}=\frac{1}{2}mv_2^2$
We plug in the known values to obtain:
$\frac{1}{2}m(\frac{40\times 1000}{3600})2-\mu_k (9.81\times 3)=\frac{1}{2}m(0)^2$
This simplifies to:
$\mu_k=2.097$
Now if the speed of the truck is $80Km/h$, then
$\frac{1}{2}(\frac{80\times 1000}{3600})^2-(2.097\times 9.81\times d)=\frac{1}{2}(0)^2$
This simplifies to:
$d=12m$
