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July 24, 2014

Homework Help: Physics

Posted by Amy on Sunday, November 12, 2006 at 11:55am.

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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