The 2050kg cable car shown in the figure descends a 200-m-high hill. In addition

Question

The 2050kg 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 1780kg counterweight up the other side of the hill. The rolling friction of both the cable car and the counterweight are negligible.

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

The 2050kg 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 1780kg counterweight up the other side of the hill. The rolling friction of both the cable car and the counterweight are negligible. a) How much braking force does the cable car need to descend at constant speed? 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

gravitational pull force along the direction of motion

for car = mgsin(30) = 10045 N

for pully = Mgsin20 = 5966.2 N

a) so for constant speed , the difference between these two forces should be equal to braking force

braking force = 10045 - 5966.2

braking force =4078.8 N

b) if brake fails

acceleration of car = (gravitational pull force of car - gravitational pull force of counter weight ) / (mass of car+mass of pully)

=4078.8/(2050+1780)

=1.065 m/sec2

distance to travel = 200/sin(30)

V2 = U2 + 2as

V2 = 0 + 2* 1.065*200/sin(30) = 851.89

V = 29.18 m/sec

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