How Is Work Calculated for a Constantly Moving Elevator?

AI Thread Summary
The discussion focuses on calculating work done by the tension in a cable and the weight of a moving elevator. For the tension, the work is calculated using the formula w = Fcos(θ)s, where θ is 0 degrees, resulting in positive work. Conversely, the weight of the elevator involves a negative work calculation since the angle is 180 degrees, indicating the force acts in the opposite direction. Participants express confusion about determining the force (F) and the angle (θ) in each scenario. Clarification on these calculations is needed for accurate work determination.
pookisantoki
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A cable lifts a 1260-kg elevator at a constant velocity for a distance of 42.3 m. What is the work done by (a) the tension in the cable and (b) the elevator's weight?

For part a.) w=(Fcos(theta))s
w=(fcos(0))s
=fs and s= 42.3

part b.) Fcos(theta))s
w=(fcos(180)s
=-fs and s= 42.3
BUt how do i find F? and is this the right set up??
 
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pookisantoki said:
BUt how do i find F? and is this the right set up??
How did you find theta in each case? In order for you to decide what is theta, should should have had an idea what was the F.
 

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