A cable lowers a 1400 kg elevator so that the elevator\'s speed increases from z
ID: 1472099 • Letter: A
Question
A cable lowers a 1400 kg elevator so that the elevator's speed increases from zero to 4.0 m/s in a vertical distance of 5.7 m. What is the force that the cable exerts on the elevator while lowering it? (Assume that the system is the elevator and Earth.) Express your answer to two significant figures and include the appropriate unite. What it the force that the cable exerts on the elevator while lowering it (Assume that toe system is the elevator only.) Express your answer to two significant figures and include the appropriate units.Explanation / Answer
Let’s use the following equation to determine the elevator’s downward acceleration.
Vfinal2 = VInitial2 + 2*a*d, VInitial = 0
4^2 = 2 * a * 5.7
a = 16 ÷ 11.4
This is approximately 1.403 m/s2. Since the elevator is accelerating as it moves downward, the net force is equal to its weight minus the tension in the cable.
Weight = 1400 * 9.8 = 13,720 N
13,720 – T = 1400 * (16 ÷ 11.4)
T = 13,720 – 1400 * (16 ÷ 11.4)
This is approximately 11755.08 N.
The force is the same in both systems. The elevator’s weight is equal to the gravitational force on the elevator.
Fg = G * M * m ÷ r^2
G = 6.67 * 10^-11
M = 5.98 * 10^24
r = 6.38 * 10^6
Fg = 6.67 * 10^-11 * 5.98 * 10^24 * 1400 ÷ (6.38 * 10^6)^2 = 10,778.99687 N
This is approximately 10,780 N.
Calculation might be faulty, logic is correct.
Hope this helps. :)
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