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## Homework Statement

An empty elevator has a mass of 722 kg. It moves between floors at a maximum speed of 6.00 m/s. The elevator is stopped on the 20th floor of the building when someone pushes the call button in the lobby (the first floor).

Assuming that it takes 15.25 meters for the elevator to reach its maximum speed, and assuming constant acceleration, calculate the tension in the cable as the elevator car begins to descend. Take the acceleration due to gravity to be 9.81 m/s2. Remember to include units with your answer.

## Homework Equations

Kinematics:

y = V

_{o}t + (1/2)at

^{2}

V

_{f}= V

_{o}+ at

V

_{f}

^{2}= V

_{o}

^{2}+ 2ay

y = t * [ (V

_{o}+ V

_{f}) / 2 ]

Newton's Second Law

F = ma

I read somewhere that tension would be Ft = m(a+g). Does this seem right?

## The Attempt at a Solution

Using kinematics I find the acceleration from which the elevator is at rest, to the point it reaches 6 m/s.

I need to solve for t to replace in another kinematic equation. I use the distance equation for it. (I am using y+ in from top to bottom).

15.25 = .5at^2

30.5 = at^2

sqrt(30.5/a)=t

I plug that into V

_{f}= V

_{o}+ at

6 m/s = a * sqrt(30.5m/a)

36 m^2/s^2 = a^2 * 30.5m/a

36 m^2/s^2 = a * 30.5m

(36 m^2/s^2) / 30.5 m = a = 1.18 m/s^2

Then I plug that into

Ft = m(a+g)

Ft = 722kg(1.18 m/s^2 + 9.81 m/s^2) = 7934.78 N

I am really bad at tension problems and want to double check that I got this correct. Thanks :)