The 1930 kg cable car shown in the figure descends a 200-m-high hill. In additio

Question

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 1880 kg counterweight up the other side of the hill. The rolling friction of both the cable car and the counterweight are negligible.
(Figure 1)

Part A

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

Express your answer with the appropriate units.

Part B

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?

Express your answer with the appropriate units.

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Figure 1 of 1

Explanation / Answer

Part A:

Here, the component of the car's weight causing it to move downhill = 1930*g*sin30

The component of the counter weight's weight causing it to resist moving uphill

               = 1880*g*sin20

So, the breaking force needed the cable car to descend at constant speed

= 1930*g*sin30 - 1880*g*sin20

= 9466.65 – 6307.81 = 3158.84 N

Part B:

Then the net force is 3158.84 N to the left (along the slope).

So, the acceleration is a = F / m = 3158.84 N / (1930+1880) kg = 0.83 m/s²
Now, the length of the run is s = 200m / sin30º = 400 m

use the expression -
v² = u² + 2as = 0 + 2 * 0.83 m/s² * 400m = 664 m²/s²
=> v = 25.8 m/s

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