## Engineering Mechanics: Statics & Dynamics (14th Edition)

$d=12m$
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: $d=12m$