The 1910 kg cable car shown in the figure descends a 200-m-high hill. In additio

Question

The 1910 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 1740 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 ,

part A) for the braking force

braking force = difference in forces on both sides

braking force = 1910 * g * sin(30 degree) - 1740 * g * sin(20 degree)

braking force = 3426 N

the braking force the cable car need is 3426 N

part B)

let the speed of the car is v at the bottom

Using conservation of energy

0.50 * (1910 + 1740) * v^2 = (1910) * 9.8 * 200 - 1740 * 9.8 * 200

solving for v

v = 13.5 m/s

the speed of car at the bottom is 13.5 m/s

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