Work done to lift an object

[Mr Davis 97]

I am a little confused about the work do the to lift an object. I know that in lifting an object a net force must be applied in order to increase its velocity, but when the object stops, gravity applies an equal but opposite force, which means that in lifting an object no net work on the object is done (work kinetic-energy theorem). Thus, how does it have gravitational potential energy when it's at the the top if no net work was done to "store" that energy in the gravitational field?

 

[DEvens]

You are wrong to say no net work is done. You lifted the object by pushing it with its own weight through a distance. Work is force times distance.

You can get that work back by letting the weight of the object push something else.

Here one should cue up the video of a teeter-totter.

 

[Mr Davis 97]

You are wrong to say no net work is done. You lifted the object by pushing it with its own weight through a distance. Work is force times distance.

You can get that work back by letting the weight of the object push something else.

Here one should cue up the video of a teeter-totter.

But using the work-kinetic energy theorem, which says that net work is equal to change in kinetic energy, isn't it so that since the initial and final velocities of an object in lifting it is zero that there is no net work done since there is no change in kinetic energy? I know that I personally did work, but no net work on the object is done because of gravity, right?

 

[DEvens]

You are wrong to say the net work is equal to the change in kinetic energy. It is not.

 

[Physicist97]

You are wrong to say the net work is equal to the change in kinetic energy. It is not.

To the best of my knowledge the net work is the change in kinetic energy. When you lift an object at constant velocity there is the gravitational force pushing it down, and a normal force pushing up. Both these forces do an equal amount of work, one negative work, the other positive. These to quantities cancel out giving you a net work of zero. You can talk about the work done by gravity, or the normal force, and that work on it's own is non-zero, but the net work is zero.

 

[SteamKing]

To the best of my knowledge the net work is the change in kinetic energy. When you lift an object at constant velocity there is the gravitational force pushing it down, and a normal force pushing up. Both these forces do an equal amount of work, one negative work, the other positive. These to quantities cancel out giving you a net work of zero. You can talk about the work done by gravity, or the normal force, and that work on it's own is non-zero, but the net work is zero.

What about change in potential energy?

If your object is sitting on the ground, it has zero potential energy. You pick the object off the ground and put it on a high shelf. The potential energy of the object is no longer zero. If you knock the object off the shelf onto your toe, you'll realize this very quickly.

 

[Mr Davis 97]

I am still confused... Could someone explain what is happening in terms of chemical, potential, and kinetic energy as a person lifts an object?

 

[nasu]

You are wrong to say the net work is equal to the change in kinetic energy. It is not.

This may be misleading unless you explain what you mean.
The work-energy theorem states that the work done by the net force is equal to the change in KE.
Do you mean that the work done by the net force is not the same as "net work"?

If the kinetic energy is the same in final and initial state, the net work is zero. The final state may have increased PE but this is related to the work done by the force associated with the PE (in this case gravity) only and not to the net work.

 

[Mr Davis 97]

If the kinetic energy is the same in final and initial state, the net work is zero. The final state may have increased PE but this is related to the work done by the force associated with the PE (in this case gravity) only and not to the net work.

So does this mean that whenever a conservative force does negative work, energy is "stored" as potential energy?

 

[Physicist97]

Net work does not tell you if there was a change in potential energy, it tells you about a change in kinetic energy. The work done by a conservative force is equal to the negative of the change in potential energy. That means even if the net work is zero, since gravity did work there was a change in gravitational potential.

 

[Doc Al]

But using the work-kinetic energy theorem, which says that net work is equal to change in kinetic energy, isn't it so that since the initial and final velocities of an object in lifting it is zero that there is no net work done since there is no change in kinetic energy? I know that I personally did work, but no net work on the object is done because of gravity, right?

That is correct. No net work is done on the object.

You are wrong to say the net work is equal to the change in kinetic energy. It is not.

No, that's true. But you must include all forces acting on the object.

 

[Qwertywerty]

I am a little confused about the work do the to lift an object. I know that in lifting an object a net force must be applied in order to increase its velocity, but when the object stops, gravity applies an equal but opposite force, which means that in lifting an object no net work on the object is done (work kinetic-energy theorem). Thus, how does it have gravitational potential energy when it's at the the top if no net work was done to "store" that energy in the gravitational field?

No net work is done , in the sense that the lifting force does +ve , and gravity , -ve work , with net zero , as stated correctly using the work - energy theorem .

As the object is lifted , gravity does negative work . It is a conservative force , and when it does negative work , it's potential energy increases . For conservative forces , ΔU_{conservative F} = -ΔW_{conservative F} . Note that this isn't dependant on the net work done , but simply on the work that that force does . Does this remove the arisen ambiguity ?

Hope this helps ,
Qwertywerty .

 

[coreluccio]

I am a little confused about the work do the to lift an object. I know that in lifting an object a net force must be applied in order to increase its velocity, but when the object stops, gravity applies an equal but opposite force, which means that in lifting an object no net work on the object is done (work kinetic-energy theorem). Thus, how does it have gravitational potential energy when it's at the the top if no net work was done to "store" that energy in the gravitational field?

In lifting an object, net work is certainly done (in general). The integral of the dot product of the net force on the object and its position vector gives the work done on the object as it is displaced. In a gravitational field, neglecting non-conservative forces, this will be exactly equal to its gravitational potential energy if the object was lifting from an initial position of rest to a final position of rest.

 

[Qwertywerty]

In lifting an object, net work is certainly done (in general). The integral of the dot product of the net force on the object and its position vector gives the work done on the object as it is displaced. In a gravitational field, neglecting non-conservative forces, this will be exactly equal to its gravitational potential energy if the object was lifting from an initial position of rest to a final position of rest.

No , as has been repeated so many times before , no net work is done .

 

[Doc Al]

In lifting an object, net work is certainly done (in general). The integral of the dot product of the net force on the object and its position vector gives the work done on the object as it is displaced. In a gravitational field, neglecting non-conservative forces, this will be exactly equal to its gravitational potential energy if the object was lifting from an initial position of rest to a final position of rest.

"Net work" means the work done by all forces, including gravity. If you neglect gravity, you are not computing net work.

 

[William White]

So does this mean that whenever a conservative force does negative work, energy is "stored" as potential energy?

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[Paul Janson]

You are saying that the work energy theorem states the Change in kinetic energy equals work done on the object. But that is an over simplification of the work energy principle which can only applied to horizontal motion.
But the real work energy principle is. The change in mechanical energy of the object is equal to the work done on the object. Mechanical energy is potential + kinetic. When lifting object at the lowest point M. E =0
at the top M. E = mgh+1/2m*0=mgh
Change In mechanical energy =work done on the object
M. Etop - M. Ebottom =mgh-0=mgh

 

[Doc Al]

You are saying that the work energy theorem states the Change in kinetic energy equals work done on the object.

Yes. As long as its the work done by all forces.

But that is an over simplification of the work energy principle which can only applied to horizontal motion.

Nope. It works just fine in general.

But the real work energy principle is. The change in mechanical energy of the object is equal to the work done on the object. Mechanical energy is potential + kinetic. When lifting object at the lowest point M. E =0
at the top M. E = mgh+1/2m*0=mgh
Change In mechanical energy =work done on the object
M. Etop - M. Ebottom =mgh-0=mgh

You are confusing the "work-energy theorem", which is really a consequence of Newton's 2nd law, with the conservation of mechanical energy. That itself is a simplification, of course.

There are some subtleties in the "work-energy theorem", but they haven't appeared in this thread yet.

 

[William White]

You are saying that the work energy theorem states the Change in kinetic energy equals work done on the object. But that is an over simplification of the work energy principle which can only applied to horizontal motion.

surely horizontal motion just depends upon what convenient set of axis you are using (we generally choose -y to mean towards the centre of the earth, but it need not be)

 

[CWatters]

The work energy theorem says that the work done by the net force on an object is equal to the change in its kinetic energy;

Wnet =ΔKE

If you take an object from rest on the ground and put it on a shelf at height h then the change in kinetic energy will be zero ΔKE=0 so

Wnet = 0

However the work done by the net force is the sum of the work done by you, and the work done by gravity.

Wnet = Wyou + Wgravity

Put these together and you get

Wyou + Wgravity = 0
or
Wyou = - Wgravity

 

[Paul Janson]

I can't understand because work is a scaler quantity and how can it be a net. I mean resultant. Only vector have resultant.
1/2m(v²-u² ) =f*s.
This is the work energy theorem. Then how it will work for problems involving elastic potential energy. Please explain me and clear my doubts

 

[Doc Al]

I can't understand because work is a scaler quantity and how can it be a net. I mean resultant. Only vector have resultant.

Net is used in the sense of "total". Net work is the total work done by all forces (including gravity).

This is the work energy theorem. Then how it will work for problems involving elastic potential energy.

Sometimes the work energy theorem is not the right tool for the job. If you have a specific problem in mind, ask about it and we'll see what the work energy theorem has to say about it.

 

[Paul Janson]

Thank you