Should work done be written as negative value?

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Discussion Overview

The discussion revolves around the concept of work done in physics, particularly whether work should be expressed as a negative value in certain scenarios. Participants explore examples involving lifting objects and the implications of gravitational force on work calculations, addressing both theoretical and practical aspects.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that work done by gravity is negative when an object is lifted against the gravitational force, as in the case of lifting a flowerpot.
  • Others argue that work is a scalar quantity and should not have a sign, suggesting that the negative sign is not necessary in the final answer.
  • One participant explains that the sign of work depends on the angle between the force and displacement, noting that if the force opposes the motion, the work is negative.
  • Another participant clarifies that the work done by the person lifting an object is positive, while the work done by gravity is negative, leading to a net work of zero if the object returns to its original position.
  • Some participants discuss the relationship between conservative forces and work, indicating that work done by conservative forces can be expressed with a negative sign in certain contexts.
  • There is confusion regarding the relationship between work and kinetic energy, with participants questioning why both are scalar quantities yet appear to have directional implications in certain contexts.

Areas of Agreement / Disagreement

Participants express differing views on whether work should be represented as a negative value, with some supporting the idea while others contest it. The discussion remains unresolved, with multiple competing perspectives on the interpretation of work done in physics.

Contextual Notes

Participants highlight the importance of understanding the definitions of work and energy, as well as the role of conservative and non-conservative forces in determining the sign of work. There are references to specific examples that may depend on the interpretation of the problem statements.

corala
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For example:

i) Someone lifts 2kg flowerpot 1m vertically upward. The work done by the gravity is -19.62 J?

Then, if
ii) A 3kg box on the floor is lifted vertically up onto the head and then lowered down to the shoulder. The shoulder is 150cm while the top of the head is 175cm measured from the floor. The work done is -44.2 J right? due to a = -9.81 as the motion of vertically upward is opposes the motion of gravitational force right? But it is written as 44.2 J in the answer sheet.

Please someone explain to me further. I'm confused between - and + value of work done. Is the - sign just indicates the opposite motion and don't need to be written as the answer or vice versa?
 
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Hey corala,

Work is NOT a vector (hence no direction). W = f * s. Although force is a vector quantity, work isn't (i.e. it's a scalar), and hence it can't have direction. Work isn't a vector quantity because it is ENERGY (hence the unit Joules - J), and energy is not a vector. It might be hard to understand, but just remember that work is energy and energy is a scalar. So that means that there should be no sign on you answers for work.

Hope this helps.
 
Work is the dot product of the force and displacement, therefore the sign of the work depends on the angle between the force and the direction of movement, or more precisely, the sign of the cosine of that angle.

If the force is in the same direction as the motion, then the angle is zero, the cosine of zero is one, and the work done is positive. If it's in the opposite direction, the cosine of 180 is -1, and the work is negative.

So the work done by gravity on an object moving upward is negative, as you said. If the answer to a problem about lifting an object to your shoulder is positive, then either it's a typo, or you didn't read the problem correctly, and it's asking for the work you do by lifting it, rather than the work gravity does in opposing it.
 
Concur with brocks; kirollos, you should read this answer carefully.
 
It depends on the force that is acting on the object.

Some conservative forces: gravitational, spring, electric.

The work done is by a conservative force so: W=F•d=-(change)Ug. or Work=- force•d.

the work donde by a non conservative force: W=F•d.

don't forget to multiply by cos(thetha) if necessary. I think the first example thetha=180
cos(180)=-1.

hope this help :)
 
I'll just add that yes, brocks gave a proper answer.

corala said:
For example:

i) Someone lifts 2kg flowerpot 1m vertically upward. The work done by the gravity is -19.62 J?
Yes. The force of gravity is downward, while the displacement is upward, in the opposite direction as the force of gravity. So the work done by gravity is negative here.
Then, if
ii) A 3kg box on the floor is lifted vertically up onto the head and then lowered down to the shoulder. The shoulder is 150cm while the top of the head is 175cm measured from the floor. The work done is -44.2 J right? due to a = -9.81 as the motion of vertically upward is opposes the motion of gravitational force right? But it is written as 44.2 J in the answer sheet.
This time we are referring to the force exerted by the person lifting the object, which is upward. The object's displacement is in the same direction, upward, so the work done is positive.

However, the work done by gravity is negative, and the net (or total) work done is zero if the object is motionless both before and after the process of lifting it.
 
Redbelly98: Correct me if I'm wrong, but i think that the first problem the work would be positive. Since in the example the work done is by a conservative force (gravity).

so it would be W=-F•d•cos180 this equal the work to be positive.

In the second example i understand that the work is positive.
 
Mapes said:
Concur with brocks; kirollos, you should read this answer carefully.

yea sorry guys, forget about my answer, it was wrong.

Now i have another question though. If work done is in joules and kinetic energy is in joules, then how come kinetic energy has no direction and work does?
 
Great question: Neither of them have direction since they are a scalar.

The negative only tells information about the object. For example: If you have an object moving at a initial speed s1 and its traveling along the x axis. Including that there is air resistance the object will slow down so its final velocity s2 will be less than s1.

Then its kinetic energy will be negative. Kf - Ki= .5[m(s2^2)-m(s1^2)], since the term {.5m(s2^2)} is less energy.
 
  • #10
boyongo said:
Redbelly98: Correct me if I'm wrong, but i think that the first problem the work would be positive. Since in the example the work done is by a conservative force (gravity).

so it would be W=-F•d•cos180 this equal the work to be positive.
No. There is no minus sign in the formula W=F·d·cosθ, regardless of what provides the force. If a person were not physically lifting the object, so that gravity were the only force on the object, it would (1) need to have some initial upward velocity in order to rise, and (2) would slow down as it rises, losing kinetic energy. A loss of kinetic energy is indicative of negative work, in accordance with the Work Energy Theorem :

Wnet = ΔKE

where Wnet = Fnet·d·cosθ
 
Last edited:
  • #11
RedBelly98: Thanks for clarifying this. I got confused with the relation between conservative forces and potential energy.

I have one last question. How can you start a thread?
 
  • #12
No problem.
boyongo said:
I have one last question. How can you start a thread?
Go to my profile page, by clicking on "Redbelly98" above my avatar. Then look for the Visitor Message dated May 8 of this year, which explains how to start a new thread.
 
  • #13
Thanks for your help Redbelly98. I was able make my first thread. Feel free to check it out.
 

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