Work at constant velocity conundrum

[FistLength]

Homework Statement

Lifting an object from point A to point B with a tension T in the string at constant velocity. What is the work done on the object?

Homework Equations

Work=Force x distance

The Attempt at a Solution

Net force = T - mg
Net F = 0 because acceleration = 0
T = mg

W = F * D

But net force is 0 because it is at constant velocity. I understand that the tension T does work =mgH
and gravity does work = -mgh
so net W= 0
but it gains potential energy...even though no work was done on it? How is this possible?

Here is the idea
http://www.batesville.k12.in.us/physics/phynet/mechanics/energy/lifting_a_book.htm

 

[nasu]

How do you define potential energy?
The change in potential energy is due to the work done by the tension in the string.

 

[FistLength]

potential energy = mass*gravity*height
The work done by the tension is equally opposed by gravity, and net work on the block is 0, yet it gains potential energy.

Say instead of vertical we moved the block horizontal on a table at constant velocity. We would say that there is no change in energy because work on the block is 0. But when we move it against gravity at constant velocity, there is a change in energy, namely change in potential energy. This doesn't make intuitive sense to me.

I am pretty lay when it comes to physics so please excuse my ignorance

 

[tiny-tim]

Hi FistLength! Welcome to PF!

potential energy = mass*gravity*height
The work done by the tension is equally opposed by gravity, and net work on the block is 0, yet it gains potential energy.

Say instead of vertical we moved the block horizontal on a table at constant velocity. We would say that there is no change in energy because work on the block is 0. But when we move it against gravity at constant velocity, there is a change in energy, namely change in potential energy. This doesn't make intuitive sense to me.

Books tend not to discuss whether to put gravity on the LHS or the RHS of work done = change in mechanical energy.

If you regard gravity as a force, then it goes on the LHS only (and in your example, the change in mechanical energy is then zero, since it is only the kinetic energy: there is no potential energy).

But if you regard gravity as an energy field, then it goes on the RHS only (and in your example, the only work done is that done by the tension).

 

[nasu]

potential energy = mass*gravity*height

No, this is not the definition but an expression for change in PE for a specific case.

The potential energy associated with a force is given by
ΔPE=-W
where W is the work of that force.
So the gravitational potential energy is
ΔPE=-W_gravity

The rest, as tiny-tim said.
Either use PE or the work done by gravity but not both.

 

[FistLength]

Hi FistLength! Welcome to PF!


Books tend not to discuss whether to put gravity on the LHS or the RHS of work done = change in mechanical energy.

If you regard gravity as a force, then it goes on the LHS only (and in your example, the change in mechanical energy is then zero, since it is only the kinetic energy: there is no potential energy).

But if you regard gravity as an energy field, then it goes on the RHS only (and in your example, the only work done is that done by the tension).

Ok I am kinda making sense of this. I don't understand the left hand side/right hand side of what you are saying. Is it possible to write a simple equation for me to see what you are talking about for both instances? Or another way to explain?

 

[FistLength]

No, this is not the definition but an expression for change in PE for a specific case.

The potential energy associated with a force is given by
ΔPE=-W
where W is the work of that force.
So the gravitational potential energy is
ΔPE=-W_gravity

The rest, as tiny-tim said.
Either use PE or the work done by gravity but not both.

Why can't you look at work done by the string and work done by gravity in the same way you would for the example of friction vs tension acting on a block moving horizontally at a constant velocity. In the latter you say add the net work, in the former we are supposed to only look at one? Why? Maybe I did not understand the explanation (likely)

 

[tiny-tim]

Hi FistLength!

I don't understand the left hand side/right hand side of what you are saying. Is it possible to write a simple equation for me to see what you are talking about for both instances?

If something falls a distance h from speed u to speed v,

then we can write the work-energy equation work done = change in mechanical energy (W = ∆E) as:

mg*h = ∆KE = 1/2 m (v² - u²)

or

0 = ∆KE + ∆PE = 1/2 m (v² - u²) - mgh

in the first equation, gravity is a force and features in the work done (LHS); in the second, it is not a force, but features as potential energy (RHS)

as nasu says …

Either use PE or the work done by gravity but not both.

 

[FistLength]

Hi FistLength!


If something falls a distance h from speed u to speed v,

then we can write the work-energy equation work done = change in mechanical energy (W = ∆E) as:

mg*h = ∆KE = 1/2 m (v² - u²)

or

0 = ∆KE + ∆PE = 1/2 m (v² - u²) - mgh

in the first equation, gravity is a force and features in the work done (LHS); in the second, it is not a force, but features as potential energy (RHS)

as nasu says …

Thanks