The 1970 kg cable car shown in the figure descends a 200-m-high hill. In additio

Question

The 1970 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 1860 kg counterweight up the other side of the hill. The rolling friction of both the cable car and the counterweight are negligible.

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

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?

Explanation / Answer

here,

The component of gravity that pulls the car along the slope is
F(slope) = F(g) * sin(theta)
F(slope) = 1970 kg * 9.81m/s^2 * sin(30degrees)

F(slope) = 9653 N

Meanwhile the retarding force of the counterweight is:
F(g) * sin(theta)
1860kg * 9.81m/s^2 * sin(20degree)

=> 6234 N

So the braking force must be 9653 N - 6234 N => 3419 N

b)

These unbalanced forces have to accelerate the entire system (1970 + 1860)kg.

v^2 = 2 a x

v^2 = 2 (F/m) x

v^2 = 2 (3419 N / (1970 + 1860)kg) * x

v^2 = 2 * 0.8913 * (200 /sin(30))

v = 26.7 m/s

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