1.An elevator in a tall building is allowed to reach a maximum speed of 3.5 m/s
ID: 1579336 • Letter: 1
Question
1.An elevator in a tall building is allowed to reach a maximum speed of 3.5 m/s going down. What must the tension be in the cable to stop this elevator over a distance of 2.5 m if the elevator has a mass of 1330 kg including occupants?
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2.
The astronaut orbiting the Earth in Figure P4.32 is preparing to dock with a Westar VI satellite. The satellite is in a circular orbit 950 km above the Earth's surface, where the free-fall acceleration is somewhat reduced. Assume the free-fall acceleration at the surface of the Earth is 9.81 m/s2 and the Earth's radius is 6400 km.
The astronaut orbiting the Earth in Figure P4.32 is preparing to dock with a Westar VI satellite. The satellite is in a circular orbit 950 krn above the Earth' surface, where the free-fall acceleration is somewhat reduced. Assune the free-fall acceleration at the surface of the Earth is 9.81 m/s and the Earth's radius is 6400 km. arth's ration at What is the frae-fall accaleration at the altitude of the satellite? 821 What is the speed of the satallita? x m/s2 375 X kmis What is the period of the satellitee. the time required to complete one orbit around te arth)? min Figure P4.32Explanation / Answer
1. vf^2 - vi^2 = 2 a d
0^2 - 3.5^2 = 2(a) (-2.5)
a = 2.45 m/s^2
Applying Fnet = T - m g = m a
T = 1330 (9.8 + 2.45)
T = 16292.5 N
2. a = G M / r^2
a r^2 = constant
(9.81) (6400)^2 = (a) (6400 + 950)^2
a = 7.44 m/s^2 .............Ans
a = v^2 / r
v = sqrt(7.44 x 7350 x 10^3) = 7395 m/s
Or 7.4 km/s .....Ans
T = 2 pi r / v = 6245 sec Or 104 min ....Ans
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