Basic Thermodynamics ~ J. Pohl © (1/19/14) www.thermospokenhere.com (C256a)   (C256)

Gravity Work-Rate:

A physics student who analyzed a physical situation in two manners and obtained contradictory results requests an explanation. The problem is stated below.
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I have a question about raising something at constant velocity. Say a block is being raised at constant velocity by a tension in a cord against the force of gravity. Is there any work done on the block?

(1) My first answer was "no" since Work = Force*Distance, Force = Mass*Acceleration, and if velocity is constant then acceleration is zero and force equals zero. So then work would be zero?

(2) However work also comprises mgΔH, and if the height is changing then there is an increase in potential energy. So my second answer is yes there is work. Please resolve this contradiction.

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Explanation:

The student's statement is unclear. Work is not germane for this "steady-continuing" event. "Work-rate" is the proper work-related form. Work (and work-rate) as ideas arrived to engineering physics about the time Watt and Newcomen developed the first steam engines. The student does not define the system for either analysis. What system, the reader must deduce.

Conclusion 1: The student is correct. The work-rate of the Block is zero as a sum of two equal and opposite work-rates in action.

Conclusion 2. The student, by mention of potential energy, is using "Block and Earth" as the system. A work-rate occurs, it is the tension of the cord times its vertical velocity. This work-rate equals the rate of increase of potential energy of the Earth/Body system.

Solution 1: For the first analysis, the student uses the Block as system. Also he makes the classic algebra-based-physics mistake. His use of "F" in Newton's 2nd Law causes him to overlook a force. One should use ΣF in the 2nd Law, as Newton intended.
So our task: Block as system, multiply Newton's 2nd Law by Block velocity. Obtain relation for work-rate.

Newton's 2nd Law addresses momentum of the Block in our "0Z" coordinate system. Best ALWAYS to use ΣF. 01 (1)
Mass of the Block and its vertical velocity are constant so left-of-equality equals zero. 02 (2)
Expand ΣF to establish gravity force as equal and opposite to the cord force. This is obvious but needed later. 03 (3)
Return to Eqn-2, multiply the equation by the vertical velocity of the Block. 04 (4)
Now expand the sum of forces.                         5 (5)
Write the forces as magnitudes and directions. 6 (6)
The vector multiplications determine the signs: 7 (7)
Force times velocity equals work-rate. The sum of these work-rates is zero. 8 (8)

Conclusion 1: The student is correct. The work-rate of the Block is zero as a sum of two equal and opposite work-rates in action.

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Solution 2.  The student's statement "work equals mgoH" is not applicable from the aspect work. Work-rate as "work-rate = d(mgoH)/dt" is relevant. The mention of increasing potential energy reveals the system to be the Block and Earth taken together. This development below addresses the Block initially with Earth participating via gravity force. Block and Earth are brought together near the end.

Newton's 2nd Law addresses momentum of the Block in our "0Z" coordinate system. Best ALWAYS to use ΣF. 9 (9)
In this and a great many instances, gravity force is relevant. Remove the force of gravity from the sum of forces. 10 (10)
As before, multiply Eqn-9 by the velocity of the Block. Three terms are obtained. 11 (11)
By basic definitions of calculus, the term left-of-equality Eqn-11 is altered as follows: 12 (12)
Right-of-equality, the first term changes as follows: 13 (13)

Since zo is a constant, dzo/dt = 0. The term is "(z - zo)" as a convenience. Place the results, Eqn-12 and Eqn-13 into Eqn-11. Thus with a little grass-roots calculus, Newton's 2nd Law multiplied by velocity, can be represented as Eqn-14.

Important: Although Newton's 2nd Law has been transformed, the system it addressed initially has not changed. That system is the Block. The terms of Eqn-14, left to right are: A modification of momentum, the work-rate of the gravity force and the work-rate of all other forces. This form, known to many persons in the years after Newton's death, displays a new idea, energy of a BODY.

14 (14)

A Body has characteristics (position and velocity) and properties (mass, momentum for now - energy soon). Supposing the mass of a Body is known, for any event its position and velocity can be made quantitative, is observable, and measurable from an external non-intefering reference (called extrinsic, by some). The terms of Eqn-14 are specified simply in terms of velocity and elevation relative to Earth.

15 (15)

When we move the gravity work-rate term from right-of-equality to left-of-equality, the effect is change our system from "Block" to "Block and Earth."

15 (16)

As a last step we define kinetic energy as KE, and potential energy as PE. Substitute these definitions into Eqn-17.

16 (17)

We can group the terms within the differentiation.

17 (18)

The above concludes the development.

Conclusion 2. The student, by mention of potential energy, is using "Block and Earth" as the system. Consequently Eqn-16 applies. Since the vertical speed is constant, the first term is zero - Eqn-19. A work-rate occurs, it is the tension of the cord times its vertical velocity. This work-rate equals the rate of increase of potential energy of the Earth/Body system.

17 (19)