A tractor with a mass of 2,300 kg is traveling at 10 m/s. At time t = 0, the driver pulls the emergency brake and the tractor begins to slide. The properties of the relevant materials, are shown:

Material 1	Material 2	Static Friction	Kinetic Friction
Rubber	Asphalt	0.9	0.4

At t = 3, the speed (m/s) of the tractor is most nearly:

To begin, lets understand the forces acting upon the tractor as it begins to slide to a stop. The force of friction between the tires and the asphalt is what exerts a force in the opposite direction of movement that allows the tractor to deaccelerate to a stop. This force is determined by the normal force, which is equal and opposite to the force of gravity, which are also simultaneously acting upon the tractor. The force diagram is shown below:

On pg. 122 of our FE Reference Handbook, we know that the force of friction is equal to the coefficient of friction multiplied by the normal force. Let's first find the normal force by calculating the force of gravity acting on the tractor:

$$\begin{gathered} F=mg \\ F = \left(2300kg\right)(9.8\frac{m}{s^2}) \\ F = 22,540N \\ N = 22540N \end{gathered}$$

Now, let's plug this into our equation for the force of friction. In this case, we need to use the kinetic coefficient of friction between rubber and asphalt, given to us in the table, since there is motion occuring:

$$\begin{gathered} F_{friction}={\mu }_{k}N \\ {F}_{friction}=(.4)\left(22540\right) \\ {F}_{friction}=(.4)\left(22540\right) \\ {F}_{friction}=9016N \end{gathered}$$

Now that we know the forces acting on the tractor in the same axis as its motion, we can use it to solve for the acceleration:

$$\begin{gathered} F = ma \\ 9106N = \left(2300kg\right)\left(a\right) \\ a = 3.92\frac{m}{s^2} \end{gathered}$$

Since this is acting on the tractor in the opposite direction of its movement, we can think of it as a negative acceleration, or -3.92m/s^2. Finally, we can use the following equation to find the speed of the tractor at t=3:

$$\begin{gathered} V_i=V_f-\left(at\right) \\ (10m/s)=\left(V_f\right)-\left(\right(-3.92\frac{m}{s^2}\left)\right(3s\left)\right) \\ 10=V_f+11.76 \\ {V}_f=-1.76m/s \end{gathered}$$

A negative velocity is impossible since there is no force that is acting to push the tractor is the opposite direction. The negative velocity simply indicates that at t=3, this was enough time to fully bring the tractor to a complete stop. Therfore, the speed of the tractor at t=3 is most nearly 0m/s.