The 1920kg cable car shown in the figure descends a 200-m-high hill. In addition
ID: 2237641 • Letter: T
Question
The 1920kg 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 1710kg counterweight up the other side of the hill. The rolling friction of both the cable car and the counterweight are negligible. (Figure 1/ http://session.masteringphysics.com/problemAsset/1383213/3/knight_Figure_08_39.jpg) 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
force on car due to gravity = mgsin30
tension in the rope T = 1710 * g *sin20
therefore Fr = braking force
mgsin30 = Fr + 1710 * g * sin20
1920*9.81 * 0.5 = Fr + 1710*9.81*0.34
Fr=3680.17 N
b)force on the car = 3680.17
acceleration = 3680.17/1920 = 1.91 m/s
distance = 200/sin 30 = 400 m
V = (2*distance *acceleration)^0.5
=39.15 m/s
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