Extremely confusing work question

[SecretSnow]

Hi guys! Another question on work done that stumbled me.. If I push an object with a constant 3500N force at a constant velocity since the friction is 3500N as well, what is the work done if the displacement is 5m to the right?? At first look, I would say its 3500*5=17500J? Because that's work done against friction right?? Is this even the same as work done on object? However, I'm thinking that it can be 0J too. The reason being that the solution says if I exert 3900N instead, there's a net 400N so 400*5=2000N. This translates to 0J if I exert 3500N instead! Which is right? Or are both right just because the bodies are different? (Is friction in this case considered as a body? I'm treating this question having having only 1 body unless you're saying the floor is another body. If so I'm guessing the work done on the body is 17500N but the work done on the floor is -17500N and the net work done is 0J. HOWEVER, the negative work done is by the object and not by me, although this is the best explanation. What is true??) thanks!

 

[Bill Nye Tho]

If Work is defined as a force being done over a distance and your Friction Force is equal to the Pushing Force then what can you conclude about the mass?

 

[ap123]

The work done by the force on the object is 3500*5=17500J
The work done by the frictional force on the object is -3500*5=-17500J
So, the total work done on the object is 17500J - 17500J = 0J

The total work done on the object is just the sum of the works done by each force.
The total work is also equal to the work done by the resultant force on the object, which in the first case is 0N.

 

[SecretSnow]

The work done by the force on the object is 3500*5=17500J
The work done by the frictional force on the object is -3500*5=-17500J
So, the total work done on the object is 17500J - 17500J = 0J

The total work done on the object is just the sum of the works done by each force.
The total work is also equal to the work done by the resultant force on the object, which in the first case is 0N.

Meaning this is still logical, given that I've done 17500J of work using my force but the resultant force's work on the object is 0J? Then what's the significance of knowing it has done 0J of work? What's the purpose in this if it does not relate to me ultimately how much work I've done? Thanks!

 

[PhanthomJay]

Meaning this is still logical, given that I've done 17500J of work using my force but the resultant force's work on the object is 0J?

yes

Then what's the significance of knowing it has done 0J of work?

No net work implies no change in the kinetic energy of the system, that is, no change in its speed, as stated.

What's the purpose in this if it does not relate to me ultimately how much work I've done? Thanks!

If friction wasn't there , you'd only after to give it a small nudge, and then it would move by itself at constant speed, so you wouln't have to do any work at all. But since friction is there, you've got to do work on the object "against friction" to keep it moving at constant speed.

 

[SecretSnow]

Oh! I suddenly remember, the work energy theorem. If there's no friction and air resistance or external forces, does it mean that when I give it a small push ill be doing infinite work since it can travel indefinitely?

 

[ap123]

work = force * distance.
As soon as you stop exerting the force, the work will drop to zero. You are only doing work as long as the force is acting.