Question

A truck of mass 10,212 kg moving at a speed of \(61.2 \mathrm{mph}\) has lost its brakes. Fortunately, the driver finds a runaway lane, a gravel-covered incline that uses friction to stop a truck in such a situation; see the figure. In this case, the incline makes an angle of \(\theta=40.15^{\circ}\) with the horizontal, and the gravel has a coefficient of friction of 0.634 with the tires of the truck. How far along the incline \((\Delta x)\) does the truck travel before it stops?

Short answer

Answer: The truck travels 91.78 meters along the incline before it stops.

Step-by-step solution

Convert the speed to m/s

First, we need to convert the speed of the truck from miles per hour (mph) to meters per second (m/s). We know that 1 mile = 1609.34 meters and 1 hour = 3600 seconds. So, the conversion factor from mph to m/s is \(1609.34/3600\). Therefore, the speed of the truck in m/s is: \(v = 61.2 \mathrm{mph} \times \frac{1609.34 \mathrm{m/mi}}{3600 \mathrm{s/hr}} = 27.353 \mathrm{m/s}\)

Find the frictional force

Next, we need to find the frictional force acting on the truck. The frictional force is given by: \(f_{f} = \mu f_{N}\), where \(\mu\) is the coefficient of friction and \(f_{N}\) is the normal force. Since the truck is moving along the incline, the normal force will have a component perpendicular to the incline. It can be determined by: \(f_{N} = mg\cos\theta\), where \(m\) is the mass of the truck, \(g\) is the acceleration due to gravity (9.81 m/s²), and \(\theta\) is the angle of the incline. Thus, \(f_{f} = \mu mg\cos\theta = 0.634\times 10,212\,\mathrm{kg}\times 9.81\,\mathrm{m/s}^2\cos(40.15^{\circ}) = 41,304.56\,\mathrm{N}\)

Calculate the change in kinetic energy

As the truck moves up the incline, its kinetic energy decreases due to the frictional force acting on it. Eventually, the truck will come to rest, meaning its final kinetic energy will be zero. We can calculate the change in kinetic energy using the formula: \(\Delta KE = \frac{1}{2}mv^2\), where \(m\) is the mass of the truck, and \(v\) is its initial velocity. So, \(\Delta KE = \frac{1}{2} \times 10,212\, \mathrm{kg} \times (27.353\, \mathrm{m/s})^2 = 3,790,388.44\, \mathrm{J}\)

Equate work done by frictional force to change in kinetic energy

The work done by the frictional force on the truck is equal to the change in the truck's kinetic energy. The work done can be expressed as: \(W = f_{f} \times \Delta x\), where \(W\) is the work done, \(f_{f}\) is the frictional force, and \(\Delta x\) is the distance the truck travels along the incline before it stops. We already calculated \(f_{f}\) and \(\Delta KE\). To find \(\Delta x\), we can set \(W\) equal to \(\Delta KE\) and solve for \(\Delta x\): \(41,304.56\, \mathrm{N} \times \Delta x = 3,790,388.44\, \mathrm{J}\)

Calculate the distance traveled

Now, we can solve for the distance the truck travels along the incline before it stops: \(\Delta x = \frac{3,790,388.44\, \mathrm{J}}{41,304.56\, \mathrm{N}} = 91.78\, \mathrm{m}\) Therefore, the truck travels \(91.78\, \mathrm{m}\) along the incline before it stops.

Key concepts

Kinetic Energy

Kinetic energy, represented by the symbol KE or simply K, is the energy possessed by an object due to its motion. It is calculated using the equation \[\begin{equation}KE = \frac{1}{2}mv^2\end{equation}\]where m is the object's mass and v is its velocity. In our truck problem, when the truck with a mass of 10,212 kg moves at 27.353 m/s, it has a substantial amount of kinetic energy. This energy must be reduced to zero for the truck to come to a stop. Understanding kinetic energy is crucial because it helps us determine how much work must be done by friction to bring the truck to a rest on an incline.

Frictional Force

Frictional force is the resistive force that occurs when two surfaces move across each other. It is always in the opposite direction of motion and depends on the types of surfaces in contact and the normal force between them. The equation for frictional force is\[\begin{equation}f_{f} = \mu f_{N}\end{equation}\]where \(\mu\) is the coefficient of kinetic friction and \(f_{N}\) is the normal (perpendicular) force exerted by the surface. In our truck's scenario, the coefficient of friction between the gravel-covered incline and truck tires is 0.634. This force plays a key role in impeding motion and is central to stopping the truck as it ascends the incline.

Inclined Plane

An inclined plane is a flat surface tilted at an angle to the horizontal. This setup creates a situation where forces acting upon an object can be resolved into parallel and perpendicular components. The normal force on an inclined plane is reduced compared to a horizontal surface, calculated as \[\begin{equation}f_{N} = mg\cos\theta\end{equation}\]where m is the object's mass, g is the gravitational acceleration, and \theta is the inclination angle. For the problem at hand, understanding the dynamics of an inclined plane allows us to determine how the truck's weight contributes to the normal force and hence the frictional force that will ultimately stop it.

Work-Energy Principle

The work-energy principle is a vital concept in physics that states the work done by all forces acting on an object will result in a change in the object's kinetic energy. The work done, W, can be expressed by the equation\[\begin{equation}W = f_{f} \times \Delta x\end{equation}\]where \(f_{f}\) is the frictional force and \(\Delta x\) is the displacement of the object. In this exercise, employing this principle allowed us to calculate the distance the truck traveled before stopping. We equated the work done by friction, which is energy expended against frictional force across a distance, to the initial kinetic energy of the truck. This approach simplifies the process of finding the stopping distance without needing to use Newton's second law directly.