# Mass moving at a constant velocity vertically

Let's say I move a mass of 10kg up by 1m at a constant velocity. I would have done 10J of work on it, and gravity would have done -10J of work.

However, wouldn't that imply that the net work done is 0J? How could the mass have increased its height?

At first, I reasoned that this is because a greater than g acceleration was applied at the start to make the mass move up at a non-zero velocity, but if this is how it increases height, lifting the mass up to 2m would require the same amount of energy.

Am I missing something?

• "Am I missing something?", no you added the -10J work done by gravity. Why do you think this is? – Jasper Sep 12 '18 at 16:56
• Are we on a planet where g = $1 \space m/s^2$ in this scenario? – Aaron Stevens Sep 12 '18 at 20:37

The key is constant velocity

By the relationship between total work done on an object and its change in kinetic energy.

$$W_{tot}=\Delta K$$

Since the object is moving at a constant velocity, $\Delta K=0$, so we must have that $W_{tot}=0$, which is what you have expressed.

So as you can see, a $0$ net work does not mean no movement. It just means no change in kinetic energy, which means no change in velocity.

The confusion might come in with thinking about potential energy. The potential energy due to gravity is related to the work done by gravity: $$W_{grav}=-\Delta U_{grav}$$

So the potential energy has actually increased, and gravity does do negative work, but the net work is still $0$, so there is no change in velocity.

At first, I reasoned that this is because a greater than g acceleration was applied at the start to make the mass move up at a non-zero velocity

I think what you mean here is that you have to apply a force greater than gravity to get the object to start moving upwards to its eventual constant speed. This is correct. Based on the above discussion, we need a non-zero net work to change the speed of the object. This is achieved by lifting with a force larger than gravity. Then, once the object is moving, we decrease our applied force to be equal to the force of gravity. This is what allows for movement at a constant velocity (which you can argue is from $0$ net force or $0$ net work. These are the same thing in 1D motion). Once again, $0$ net work does not mean $0$ displacement, just like how $0$ net work does not mean $0$ velocity.

We can even take this a step further and think about what happens when the object comes to rest again above our head. Then we know that the net work from start to finish is $0$, since we started and ended at rest. What gives? We did work lifting the object didn't we? We did! But gravity did the same amount of negative work the whole time. Another way to look at it is that we used energy to increase the potential energy of the object. But potential energy is not essential to understanding the work done in this system.

Side note, I think you need to check your calculations on how much work is done by each force during the constant velocity movement you discuss in the question.

In your question you didn’t state whether or not the mass is brought to rest at 1 meter, or if it continues past 1 meter with the same velocity. It makes a difference because if it still has velocity at 1 meter is has kinetic energy equal to $1/2mv^2$ at 1 meter in addition to the 100 J of potential energy (correcting for g = 10 and m = 10). If the external force remains then you will gain another 100 J of potential energy when you get to 2 meters. In the following, the mass will have constant velocity until just before it reaches 1 meter when it will decelerate to zero velocity at 1 meter.

Recall that the gravitational force is a conservative force. This means if I start with a mass at rest at point 1, say with respect to the surface of the earth, and move the mass along any arbitrary path P up to point 2 and back to point 1, with any arbitrary velocity along the way, the net work done (me + gravity) will be zero. In the following scenario all motion is vertical, i.e., parallel to the force of gravity, since it should be obvious to all that no work is required (absent air friction) for horizontal motion.

I start with moving the mass from point 1 to point 2. Clearly, to accomplish this, the mass has to move, i.e., it must attain some velocity along the way. To get things going I need to apply an external upward force, $F_{ext}$, slightly greater than $mg$, to give it small upward acceleration $a$. Let’s say I do this for a brief time $dt$ and therefore over a short distance $dh$. The mass thereby attains a small velocity $v=adt$ and a small increase in kinetic energy of $½ m(adt)^2$ and a small increase in the potential energy of the mass m of $(mg) dh$. By the work-energy principle:

$$W = \Delta KE + \Delta PE$$

Since the initial KE and PE are zero, this becomes, for a differential time $dt$ and a differential increase in height of $dh$

$$dW = ½ m(adt)^2 + (mg) dh$$

We now immediately reduce our upward force so that it equals the downward gravitational force. The mass is now rising at constant velocity v and so there is no subsequent change in KE, however its potential energy keeps increasing. Re-applying the work-energy principle, realizing now that KE is constant and $\Delta KE=0$

$$dW = (mg)dh$$

We now ask ourselves, what is doing this work. Clearly, the only force that is responsible for raising the mass is the external upward force that I exert in order to perform work against gravity.

But so far we are still left with the small kinetic energy.

Therefore, prior to reaching point 2 I reduce my external upward force, $F_{ext}$, to slightly less than $mg$, to give it small acceleration downward. I do this for sufficient time to bring the mass to rest at point 2. Gravity now does a small amount of work downward resulting in a negative change in kinetic energy. The mass continues to rise in height during this period., Re-applying the work-energy principle for this differential amount of time $dt$ and increase in height $dh$:

$$dW =-½ m (adt)^2 + (mg)dh$$

The end result in going from 1 to 2 is an increase in potential energy of

$$\int_1^2 (mg)dh = (mg)h$$.

In order to return from point 2 to point 1 I reverse the above procedure, starting with an upward force slightly less than the gravitational force for a downward acceleration to attain a downward constant velocity $v$ and end with an upward force slightly more than the gravitational force to bring the mass to rest at point 1. The end result is a change in potential energy of $-mgh$. In this case, negative work is done by gravity equal to the positive work done by the external force going from 1 to 2, for a net work of 0.

Before closing, a different scenario can consist of moving the mass from 1 to 2 as above, giving it an increase in PE of $mgh$ and then releasing the mass for free fall. Gravity then converts all the potential energy to kinetic energy prior to impact. Ultimately, where does this final kinetic energy come from? It came from the work that was done by the external agent to give the mass its potential energy at point 2, and thus energy is conserved going from 1 to 2 and back to 1.

Hope this helps.