Question

You are working with a team that is designing a new roller coaster-type amusement park ride for a major theme park. You are present for the testing of the ride, in which an empty 220 kg car is sent along the entire ride. Near the end of the ride, the car is at near rest at the top of a 101 m tall track. It then enters a final section, rolling down an undulating hill to ground level. The total length of track for this final section from the top to the ground is 250 m. For the first 230 m, a constant friction force of 350 N acts from computer-controlled brakes. For the last 20 m, which is horizontal at ground level, the computer increases the friction force to a value required for the speed to be reduced to zero just as the car arrives at the point on the track at which the passengers exit. (a) Determine the required constant friction force (in N) for the last 20 m for the empty test car.

Answer 1

Answer:

The required constant friction force for the last 20 m is 6,862.8 N

Explanation:

Energy Conversion

There are several ways the energy is manifested in our physical reality. Some examples are Kinetic, Elastic, Chemical, Electric, Potential, Thermal, Mechanical, just to mention some.

The energy can be converted from one form to another by changing the conditions the objects behave. The question at hand states some types of energy that properly managed, will make the situation keep under control.

Originally, the m=220 kg car is at (near) rest at the top of a h=101 m tall track. We can assume the only energy present at that moment is the potential gravitational energy:

$E_1=mgh=220\cdot 9.8\cdot 101=217,756\ J$

For the next x1=230 m, a constant friction force Fr1=350 N is applied until it reaches ground level. This means all the potential gravitational energy was converted to speed (kinetic energy K1) and friction (thermal energy W1). Thus

$E_1=K_1+W_1$

We can compute the thermal energy lost during this part of the motion by using the constant friction force and the distance traveled:

$W_1=F_{r1}\cdot x_1=350\cdot 230=80,500\ J$

This means that the kinetic energy that remains when the car reaches ground level is

$K_1=E_1-W_1=217,756\ J-80,500\ J=137,256\ J$

We could calculate the speed at that point but it's not required or necessary. That kinetic energy is what keeps the car moving to its last section of x2=20 m where a final friction force Fr2 will be applied to completely stop it. This means all the kinetic energy will be converted to thermal energy:

$W_2=F_{r2}\cdot x_2=137,256$

Solving for Fr2

$\displaystyle F_{r2}=\frac{137,256}{20}=6,862.8\ N$

The required constant friction force for the last 20 m is 6,862.8 N