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 the 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.