# when an object is lifted (at a constant velocity) shouldn't the work done on the object be zero? [duplicate]

When I lift an object from the ground (at a constant velocity) I'm applying force on the object equal to its weight and the earth is also pulling it downwards with equal amounts of force. So if the net force on the object is zero shouldn't the WORK also be zero?...(Although the potential energy increases which indicate work done.....I'm a bit confused.)

When i lift an object from the ground (at a constant velocity) I'm applying force on the object equal to it's weight and the earth is also pulling it downwards with equal amounts of force. So if the net force on the object is zero shouldn't the WORK also be zero?

You should consider the definition of work

In physics, a force is said to do work if, when acting on a body, there is a displacement of the point of application in the direction of the force. For example, when a ball is held above the ground and then dropped, the work done on the ball as it falls is equal to the weight of the ball (a force) multiplied by the distance to the ground (a displacement)

If you apply a force to an object and it is lifted from the ground, that simply means that you have done positive work on that object, because you have displaced it and the amount of work is its weight times the displacement.

If work done were zero the object would remain on the ground

No that's not right. a work can be calculated for each force individually. you are mixing it with the equation between work and energy that says:

$$W = \Delta K \quad \text{K is kinetic energy}$$

but this work is actually the work of the overall force on the object. the work done by earth gravity force and your force is zero. But each is doing some work individually. you are doing some positive amount of work and the earth is doing it in the same amount but in negative sign.

This appears to be a very normal intuition challenge: someone has claimed the math says the world works one way, and intuition declares it works the other. Let's see if we can get intuition to agree with the so called "math."

Gravity is a sneaky thing. Even at the Quantum Mechanical level, it keeps causing problems. It will be easier to understand if we look at an analogous situation using a spring instead of gravity.

Dust off your frictionless surface and lay it on the ground. You ordered one of these from Amazon, right? Physics is so much easier once you have the proper laboratory equipment like frictionless surfaces and spherical cows. In fact, just because this is my analogy, go grab your spherical cow... we'll use it as the object we're trying to move around and do work on.*

Now lets grab a spring. We're going to hook one end of the spring to a Fixed Point™ in the middle of our frictionless surface. Hook the other end to the spherical cow. Now this situation is akin to the scenario you described with gravity working on an object, but it has been replaced with a spring so that we can more intuitively see where the work is going as we stretch things.

Now we're going to push the cow away from the center at a fixed velocity (your frictionless surface should have come with special "Sticky Shoes" so that you can do this. Never attempt to push a spherical cow on a frictionless surface without Sticky Shoes). Obviously the spring will begin to stretch. We can intuitively recognize that the spring is building up potential energy, because it would snap back if we stopped pushing. Thus we should recognize intuitively that we are doing work, but how should we think about this to make it more applicable to your original example?

One of the definitions of work is a force applied in a direction for a distance ($W=F\cdot s$, where s is the direction of movement). If the force is applied in the direction of movement, it's positive work (work done to the object by the actor). If the force is applied opposed to the movement, it's negative work (work done to the actor by the object). Now let's break apart all of the forces and see how they work:

• You are applying force $F_{you\to cow}$ on the cow in the direction of movement, so your work, $W_{you\to cow}$, is positive (you are applying work to the cow. Mooo!).

• The spring is applying a force on the cow, $F_{spring\to cow}$, and the cow is naturally applying a resulting force back, $F_{cow\to spring}$. Now the direction of these forces is important. Because the cow is effectively stretching the spring, it is doing work in the positive direction, so it is doing work ($W_{cow\to spring} >0$), and the spring is having work done to it ($W_{spring\to cow} < 0$). Initially, we're going to focus on $W_{cow\to spring}$ because it follows intuition better: you're applying work to the cow, and it's applying it to the spring. It's a lot easier to think of all of the positive works when you can -- most brains aren't wired to think of "negative work."

Now, the spherical cow is not accelerating, as per your question. This means the sum of the forces must be equal to zero, so $F_{you\to cow} = -F_{spring\to cow}$. Since I mentioned earlier it's easier to think of a long chain of works from you to the cow to the spring, lets flip the direction on the second term by dropping the negative sign and switching the order of the actor and object: $F_{you\to cow} = F_{cow\to spring}$. From this wording, we see that all of the force you apply to the cow is applied by the cow to the spring (which makes sense, since the cow isn't accelerating). This means $W_{you\to cow} = W_{cow\to spring}$. All of the work done to the cow must be done to the spring.

That work has to go somewhere, so it goes into deforming the physical structure of the spring, straining atomic bonds and what not to store the energy to be released later.

Now we can bring this back to your original question by replacing the frictionless surface and spring with an object being lifted into the air at a constant rate. You are still pushing an object away from a fixed point (the ground beneath your feet), against a conservative force (gravity, which happily snaps back when you stop pushing on it, just like the spring did). The last piece of the puzzle is "what is the equivalent of the 'physical structure of the spring deforming." The answer is that it's the "deformation" of objects being drug through a gravitational field. In this specific case, it is not too bad to think of the effects of gravity being like a big rubber sheet that tries to pull back on objects, storing up "potential energy." However you want to think about it, the work you expend lifting the object gets "stored" in gravitational potential energy, just as the work expended pushing the cow got "stored" in the spring's potential energy.

* If you've not seen this before, there is an ongoing joke in Physics education about how we love to use impossible objects because they make the math easier, but we want to use tangible examples. One of the most famous questions involved finding the trajectory of a cow through the air. At the end of the problem was a caveat, "Assume the cow is a perfect sphere."

Now, you'll note I dropped a ton of resultant forces in my example. They don't matter when you're just approaching this problem on the surface. However, they get extremely interesting when you try to expand this problem to include more complicated problems, like putting an object into orbit.

In the original example, we used some pretty fancy laboratory equipment. Fixed Points™ that can be attached to frictionless surfaces are very expensive, and Sticky Shoes do eventually wear out. What if we tried to do this experiment without them? (I'll keep the frictionless surface because it lets me model things being lifted into the air without having to pay attention to gravity itself. It keeps the model sane.)

Instead of attaching the spring to a Fixed Point™, lets just stick it on an ordinary object, like a block. Instead of using Sticky Shoes to push directly on the frictionless surface, lets push against the block yourself. Depending on if you're visualizing moving it a little bit or a lot, you can either visualize this as you're going to push off the block by extending your legs, or you may have to attach a walk-way of sorts to the block so that you can push it. It doesn't matter for the physics, just your intuitive model of what it looks like to push spherical cows around a frictionless surface.

What does matter is that all of the force your legs used to apply to the ground must now be applied to the block. Now, if you take the time to account for all of those pesky resultant forces like $F_{cow\to you}$. When you do so, you find that the sum of forces on you, the spring, and the cow are all 0 because the resultant forces cancel everything out nicely (if they didn't we'd be accelerating). The last two resultant forces that will not have been canceled out are $F_{you\to block}$ and $F_{spring\to block}$. Since the block isn't accelerating either, these two forces are equal and opposite as well.

Now this is neat because, unlike the first example, the frictionless surface is no longer having any particularly special effect. All it's doing is holding in a 2d plane to make it easier to describe what is going on in words. We've removed virtually all of the fancy laboratory equipment, and all that is left are normal objects from the real world!

So in this case, when you push the cow outwards, it stretches the spring. The block gets pushed slightly backwards, and you and the cow push slightly forward. The entire system "gains energy," just like when you stretch a huge gym rubbber band over your head and realize just how much potential energy you stored in it (and just how inadequate your muscles are for keeping it there without it snapping back on your head).

Likewise, in your real question, when you finally get around to modeling all of the resultant forces, you'll find that lifting the cow actually pushes backwards on the earth, stretching the virtual-rubber-band of gravity. This "stretching" is the potential energy of gravity.

And this is what happens when you let go, and all of the gravitational potential energy snap back.

It is zero; think of the conservation of energy. Actually, because of this law, all force's work on an object will be equal to only its change of kinetic energy (at least in the Newtonian physic), and in this case this is zero. We use the potential energies to ease up the calculations (so you don't have to calculate all times i.e. the Coulomb-force's work): they are some kind of a "reverse work" - the amount of the energy can be gained in the future. By definition: the gravitational potential energy is the work of the gravitational force while it takes the object from height $h$ compared to the zero point to the zero point. This is the opposite of the gravitational work in our case. But this can not be used when you count with the force's work since in this way you would count it twice: you can either use $$W_F +W_{\text{grav}} = 0 \quad \text{or} \quad W_F = E_\text{grav.pot}$$ but never $$W_F + W_\text{grav} =E_\text{grav.pot}$$.

Yes, the net force is zero. But that doesn't mean you are doing no work "against gravity".

That the system is in stable equilibrium, does indicate that the kinetic energy is constant. You are changing the configuration of the system- its state of position - against a field where the force is given by the gradient of potential energy ie. $$\mathbf{F} = -\nabla U$$. So, you have to do work against the field concerned so as to change the state. If the system had changed its velocity, that only meant you were doing extra work for executing the change in kinetic energy.