Work done by and against a force

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    Force Work Work done
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Discussion Overview

The discussion revolves around the concepts of work done by a force versus work done against a force, exploring their definitions, calculations, and relationships. Participants delve into examples involving gravitational force and other opposing forces, examining the nuances of these terms in various contexts.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants question the semantics of "work done by" versus "work done against" a force, seeking clarification through examples.
  • Participants discuss how to calculate work done by gravity and the work done against gravity, with references to formulas like W=FSCostheta and W=(-)mgh.
  • There is a suggestion that the work done against gravity can be viewed as the work done by a person lifting an object, equating to mgh when moving at constant speed.
  • Some participants argue that if the force exerted by a person equals the gravitational force, the net force is zero, leading to no movement, which raises questions about how work is calculated in such scenarios.
  • There are discussions about the implications of applying a force greater than mg to initiate movement and how this relates to work done against gravity versus work done to increase kinetic energy.
  • Participants explore the idea that work done against a force may not account for kinetic energy changes, emphasizing the distinction between these concepts.

Areas of Agreement / Disagreement

The discussion contains multiple competing views and remains unresolved regarding the precise definitions and calculations of work done by and against a force. Participants express differing interpretations and examples without reaching a consensus.

Contextual Notes

Limitations include varying interpretations of terminology, assumptions about forces in motion, and the conditions under which work is calculated. The discussion reflects a range of perspectives on how to approach these concepts mathematically and conceptually.

  • #31
cnh1995 said:
Consider a box placed on a smooth surface and a man is pushing it with force F. If the box moves through a distance s, work done "on the block"=F.s and is positive. If the box were moving with some velocity v and the man applied an opposing force F, the block will decelerate and stop. Here also, work is done "on the block by the man" , but F and s are in opposite direction, hence work done is negative here. Whoever applies force, does the work. If a man is moving a body up against gravity such that net force on it is 0, this means mg=Fman.
Since F(man) and s are in the same direction, work done by the man is positive and man loses muscular energy. Work done by gravity is negative since mg and s are opposite in direction. Hence, the body will gain same amount of PE. K.E. of the body will be 0 as discussed earlier( or it is unaffected, if present).
I had a question about positive and negative work, too. Could you please check it out? https://www.physicsforums.com/index.php?threads/851409/
 
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  • #32
D. Wani said:
But, as cnh1995 said, for the body to be set into motion, the initial force has to be in infinitesimally greater than mg, just at that instant.. After that, to maintain constant speed, we need to apply a force equal to (mg). Right?
Sure, if it starts from rest you must first accelerate the box.
 
  • #33
D. Wani said:
But, as cnh1995 said, for the body to be set into motion, the initial force has to be in infinitesimally greater than mg, just at that instant.
To make KE=0, the force should be infinitesimally larger. Even if the body has nonzero KE and net force is 0, KE is not affected as I've said in earlier post.
 
  • #34
D. Wani said:
Though work is only done when there is displacement of the body
And what? Work is not always exsists even in the presence of motion. It is necessary that present a non-zero projection of force in the direction of the moving. No matter what the sign is. But it does not change anything. The question is simple, and until you complicate it, you will spend the time for nothing. All the best.
 
  • #35
Can work done against gravity be negative? (In the sense that the angle between displacement and force is zero)
 
  • #36
D. Wani said:
Can work done against gravity be negative? (In the sense that the angle between displacement and force is zero)
Why not? What if you lower the box?
 
  • #37
D. Wani said:
Can work done against gravity be negative? (In the sense that the angle between displacement and force is zero)
The work must be negative aganst the whatever you want force. And it always is! In condition if the "whatever you want force" direction is accepted as possitive. You still think about things that are not actual for you. All should be done in sequence. Let you understand what is work of force against another force is. The other things after that.
 
  • #38
D. Wani said:
Can work done against gravity be negative? (In the sense that the angle between displacement and force is zero)
In my earlier man and box example, when both F and s were in the same direction, work done "by the man" was positive and hence, he lost energy. In the next example also, work done by the man was positive and that by gravity was negative, hence, man lost energy and the body gained gravitational PE. So, by convention, whoever does positive work, loses energy and whoever does negative work, gains energy. If you want to do 'negative' work 'against' gravity, you should gain energy i.e.you should impede the body's motion "due to" gravity. When you catch a free-falling ball, the ball's motion is due to gravity and you impede it, thereby doing negative work against the gravity. Hence, you(your hands actually) gain the energy lost by the ball.
 
Last edited:
  • #39
cnh1995 said:
So, by convention, whoever does positive work, loses energy and whoever does negative work,
That is thermodynamical look on a things. But in mechanical sense it absolutely does not metter what force works against to another.
$$A =F\cdot s\cdot \cos(\alpha )$$
where F and s are modules, but $$cos(\alpha)$$ determines the sign.
 
  • #40
Thank you, everyone
 

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