Physics
posted by Amy .
An elevator and its load have a combined mass of 1600 kg. Find the tension in the supporting cable when the elevator, originally moving downward at 12 m/s is brought to rest with constant acceleration in a distance of 42m.
T=w which means that T=mg. My book said that the answer is 1.8X10^4 N. But I can't get the answer. What am I doing wrong?
You must apply Newton's second Law:
F = M a,
where F is the force acting on a mass m and a is the acceleration of that mass.
Gravity acts on the elevator with a force in the downward direction of F_g = 1600 kg*9.81 m/s^2 = 15696 N
If the cable exerts a force T on the elevator in the upward direction of T, then the total force exerted on the elevator in the downward direction is:
F_{total} = 15696 N  T
The acceleration in the downward direction is then:
a = F_{total}/m = g  T/M
Now you ca deive from the given data that the lift is accelerating in the upward direction. You can write the acceleration in the downward direction as:
a = 1/2 (12 m/s)^2/(42m) = 1.714 m/s^2
So:
g  T/M = 1.714 m/s^2 >
T = 1.8*10^4 N
(Note that it is a good practise to keep more significant figures in the intermediary steps and round it off at the end).
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