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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 P4.32

Explanation / 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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