How Is Work Calculated for a Constantly Moving Elevator?

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SUMMARY

The work done by the tension in a cable lifting a 1260-kg elevator at a constant velocity over a distance of 42.3 m can be calculated using the formula w = (Fcos(theta))s, where theta is 0 degrees, resulting in w = fs. Conversely, the work done by the elevator's weight is calculated with w = (Fcos(180))s, yielding a negative value due to the opposing direction of force, where theta is 180 degrees. The key to solving these problems lies in accurately determining the force (F) exerted by the cable and the gravitational force acting on the elevator.

PREREQUISITES
  • Understanding of Newton's Second Law of Motion
  • Familiarity with the concept of work in physics
  • Knowledge of trigonometric functions, specifically cosine
  • Basic grasp of forces acting on objects in motion
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  • Study the principles of work and energy in physics
  • Learn how to calculate forces in static and dynamic systems
  • Explore the application of trigonometry in physics problems
  • Investigate the relationship between mass, weight, and gravitational force
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Students studying physics, educators teaching mechanics, and anyone interested in understanding the calculations involved in work done by forces in motion.

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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