Newton's 2nd law with an incline plane (I think help0

[Kalie]

Newton's 2nd law with an incline plane (I think...help0

The 1930 kg cable car shown in the figure descends a 200-m-high hill. In addition to its brakes, the cable car controls its speed by pulling an 1830 kg counterweight up the other side of the hill. The rolling friction of both the cable car and the counterweight are negligible.

How much braking force does the cable car need to descend at constant speed?

I solved for this and found that the braking force was 3320 N needed

One day the brakes fail just as the cable car leaves the top on its downward journey. What is the runaway car's speed at the bottom of the hill?

I have no clue how to approach this I drew out my diagrams but with every anwer I came up with I was wrong I think it was the setup

If you would mind could someone just explain the setup please?

 

[OlderDan]

You have not given enough information for us to verify your braking force result. I will assume you did that part correctly. If so, when the brakes fail there will be a net force on the cable car in the direction it moves down the hill of 3320N; this is the difference between the weight component acting down the hill and the tension in the cable connected to the counterweight. The counterweight has a net force acting up the other side of the hill in the direction the weight moves; this is the difference between the tension in the connecting cable, and the weight component acting down the hill.

Apply Newtons second law to each of the connected objects using the unknown tension and acceleration. Recognizing that both objects experience the same tension and the same magnitude of acceleration. You can solve for both of these unknowns. Then, using the acceleration and teh distance the car moves down the hill you can calculate the speed of the car at the bottom.

Alternatively, if you are familiar with conservation of energy approaches to such problems, you can calculate the changes in potential and kinetic energies of the car and the counterweight to find the answer.

 

[kyle8921]

The setup for this problem involves using Newton's Second Law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration (F=ma). In this case, the cable car is experiencing a downward force due to gravity, and an upward force due to the counterweight pulling in the opposite direction. The net force on the cable car will be equal to the difference between these two forces, which can be calculated using the equation F=mg, where m is the mass of the object and g is the acceleration due to gravity (9.8 m/s^2).

To find the braking force needed for the cable car to descend at constant speed, we need to set the net force equal to zero, since the car is not accelerating. This gives us the equation mg - mg = 0, which simplifies to 0 = 0. This means that the net force is indeed zero, and the cable car will descend at a constant speed without needing any additional braking force.

However, if the brakes fail and the cable car starts accelerating downward, the net force will no longer be equal to zero. In this case, we can use the equation F=ma again to calculate the acceleration of the cable car as it descends the hill. Once we have the acceleration, we can use the kinematic equation v^2 = u^2 + 2as to solve for the final velocity (v) at the bottom of the hill, where u is the initial velocity (which we can assume is zero) and a is the acceleration calculated from the net force acting on the cable car. This will give us the speed of the runaway cable car at the bottom of the hill.

It is important to note that this calculation assumes ideal conditions, without taking into account any external factors such as air resistance or friction. In a real-life situation, the speed of the cable car may be different due to these factors.