A truck of mass 9612 kg moving at a speed of 73.2 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 θ = 37.95° 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 (Δx) does the truck travel before it stops?
To find how far the truck travels along the incline before it stops, we need to use the principles of Newton's laws of motion.
First, let's convert the speed of the truck from mph to m/s, because SI units are typically used in physics calculations. We know that 1 mph is equal to 0.44704 m/s. So, the speed of the truck in m/s is:
73.2 mph * 0.44704 m/s = 32.7232 m/s
Next, we need to resolve the gravitational force acting on the truck into components parallel and perpendicular to the incline. The component of the gravitational force parallel to the incline will oppose the movement of the truck, while the perpendicular component will be balanced by the normal force.
The parallel component of the gravitational force can be calculated using:
F_parallel = m * g * sin(θ)
where m is the mass of the truck (9612 kg), g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of the incline (37.95°).
F_parallel = 9612 kg * 9.8 m/s^2 * sin(37.95°) = 56115.06 N
Now, we can calculate the frictional force acting on the truck using:
F_friction = coefficient of friction * F_normal
where F_normal is the normal force and can be calculated using:
F_normal = m * g * cos(θ)
F_normal = 9612 kg * 9.8 m/s^2 * cos(37.95°) = 76394.21 N
F_friction = 0.634 * 76394.21 N = 48435.16 N
Since the frictional force acts opposite to the motion of the truck, it will help to decelerate the truck until it comes to a stop. As the truck decelerates, the net force acting on it will be the difference between the parallel component of the gravitational force and the frictional force:
Net force = F_parallel - F_friction
Net force = 56115.06 N - 48435.16 N = 7679.9 N
Using Newton's second law of motion (F = m * a), where a is the acceleration, we can find the acceleration of the truck:
7679.9 N = 9612 kg * a
a = 7679.9 N / 9612 kg = 0.7996 m/s^2
Now, let's find the distance (Δx) the truck travels before it comes to a stop. We can use the equation of motion for uniform acceleration:
Δx = (v^2 - u^2) / (2a)
Where v is the final velocity (0 m/s) when the truck comes to a stop, and u is the initial velocity (32.7232 m/s). By substituting the values, we can find:
Δx = (0^2 - (32.7232 m/s)^2) / (2 * 0.7996 m/s^2)
Δx = -1073.2 m^2/s^2 / 1.5992 m/s^2
Δx = -670.24 m^2/s^2
As we can see, the square of the initial velocity is greater than the square of the final velocity, resulting in a negative value. However, distance cannot be negative, so the negative sign indicates that the truck stops at a distance of 670.24 meters along the incline.
Therefore, the truck will travel approximately 670.24 meters before it comes to a stop.