Calculating Work for a Constant Velocity Elevator

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SUMMARY

The discussion focuses on calculating work done by the tension in a cable and the weight of a 1200 kg elevator moving at a constant velocity over a distance of 35 m. The correct calculations yield 4.1 x 105 J for the tension and -4.1 x 105 J for the weight. Initial misconceptions arose from the assumption that constant velocity implies zero force and work, which is incorrect. The net force is zero, but individual forces do perform work, leading to non-zero values for both tension and weight.

PREREQUISITES
  • Understanding of Newton's laws of motion, particularly the relationship between force, mass, and acceleration.
  • Familiarity with the work-energy principle, specifically the equation W = fscos(theta).
  • Basic knowledge of vector components and their impact on work calculations.
  • Concept of net force and its implications on work done in mechanical systems.
NEXT STEPS
  • Study the implications of constant velocity on force and work in mechanical systems.
  • Learn about the work-energy theorem and its applications in physics problems.
  • Explore examples of work done by various forces in different scenarios, such as friction and tension.
  • Investigate the relationship between net force and individual forces in dynamic systems.
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Students studying physics, particularly those focusing on mechanics, as well as educators looking to clarify concepts related to work and energy in constant velocity scenarios.

hey123a
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Homework Statement



A cable lifts a 1200
kg elevator at a constant velocity for
a distance of 35 m. What is the work done by
(a)
the tension in the cable and
(b)
the elevator's weight?


Homework Equations



W = fscos(theta)

The Attempt at a Solution



okay i already know the answer to the solution and how to do it, however, when i first interpreted the question i thought the answer was zero because it says that it moves at a constant velocity. if its at a constant velocity then acceleration is zero. since f = ma, and acceleration is zero then force is equal to zero, so that means work is also equal to zero since W=fscos(theta)
however the answer to the question is actually

Solution: (a) 4.1 x 10 5J, (b) -4.1 x 10 5J

again, i know how to get the right answer but i would like to know what was wrong with my initial thinking
 
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hey123a said:

Homework Statement



A cable lifts a 1200
kg elevator at a constant velocity for
a distance of 35 m. What is the work done by
(a)
the tension in the cable and
(b)
the elevator's weight?


Homework Equations



W = fscos(theta)

The Attempt at a Solution



okay i already know the answer to the solution and how to do it, however, when i first interpreted the question i thought the answer was zero because it says that it moves at a constant velocity. if its at a constant velocity then acceleration is zero. since f = ma, and acceleration is zero then force is equal to zero, so that means work is also equal to zero since W=fscos(theta)
however the answer to the question is actually

Solution: (a) 4.1 x 10 5J, (b) -4.1 x 10 5J

again, i know how to get the right answer but i would like to know what was wrong with my initial thinking

The net force is zero.Hence net work done is zero.But the work done by the tension in the cable and the weight are non zero.

You can easily deduce this from the answers you have got.The magnitude of the work done by the two forces are equal,but that of the tension is positive while that of the weight is negative.
 
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