#### If the load is on the point of sliding forward on the bed of the slowing truck, static friction acts backward on the load with its maximum value, to give it the same acceleration as the truck:

$$\sum F_{x} = ma_{x} : − f_{s} = m_{load}a_{x}$$

$$\sum F_{y} = ma_{y} : n − m_{load}g = 0$$

Solving for the normal force and substituting into the $$x$$ equation gives:

$$-\mu_{s}m_{load}g = m_{load}a_{x}$$ or $$a_{x} = −\mu_{s}g$$

We can then use

$$v_{xf}^{2}=v_{xi}^{2}+2a_{x}(x_{f}-x_{i})$$

Which becomes

$$0=v_{xi}^{2}+2(-\mu_{s}g)(x_{f}-0)$$

(a) $$x_{f}=\frac{v_{xi}^{2}}{2\mu_{s}g}=\frac{(12.0m/s)^{2}}{2(0.500)(9.80m/s^{2})}=14.7m$$

(b) From expression $$x_{f}=\frac{v_{xi}^{2}}{2\mu_{s}g}$$

neither mass affects the answer.