Question

Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loose gravel are constructed to stop runaway trucks that have lost their brakes. The combination of a slight upward slope and a large coefficient of rolling resistance as the truck tires sink into the gravel brings the truck safely to a halt. Suppose a gravel ramp slopes upward at $6.0^o$and the coefficient of rolling friction is $0.40$. Use work and energy to find the length of a ramp that will stop a $15,000kg$ truck that enters the ramp at $35m/s\left(\approx 75mph\right)$.

Short answer

The distance a $15000kg$truck rolls on the gravel surface before the ramp stops it is$124.29m.$

Step-by-step solution

Given information

A gravel ramp goes upward at $6.0^o$and the coefficient of rolling friction is$0.4$. The truck enters the ramp at$35m/s$.

Explanation

The frictional force acting on a sliding mass m on a surface with a coefficient of kinetic friction $\mu is\mu mgh$.

The KE of the truck before entering the ramp is

$KE=\frac{1}{2}m{v}^2$

This energy is transferred to the mass and used in doing work against the friction force. Work done by friction + gain in PEis

$$\begin{gathered} W=Force\times dis\tan ce+gaininPE \\ W=\left(\mu mg\right)d+mgh=KE \\ \left(\mu mgh\cos \left(6^o\right)\right)d+mgh=\frac{1}{2}m{v}^2 \\ \left(\mu mg\cos \left(6^o\right)\right)d+mgd\sin \left(6^o\right)=\frac{1}{2}m{v}^2 \\ d=\frac{v^2}{2g\left[\mu \cos \left(6^o\right)+\sin \left(6^o\right)\right]} \\ d=\frac{{35}^2}{2\times 9.81\left(0.4\cos \left(6^o\right)+\sin \left(6^o\right)\right)} \\ d=124.29 \end{gathered}$$

The energy is conserved. This energy of the truck is used up in doing work against the friction force and gaining PE.Thus, the truck finally comes to a halt.