Why don't we take kinetic energy into account when finding the change in potential energy?

Question

An engine draws water from depth of $10 \;\text{m}$ with constant speed $2 \;\text{m/s}$ at rate of $1 \;\text{kg/s}$. What is the power of the engine?

While solving the question, I found that we involved both kinetic energy and potential energy. However, in general, when we are solving questions like

Find the change in energy when an object moves upwards by a given height

why do we only talk about potential energy or use the simple formula($\mathbf{F} \cdot \mathbf{s}$) and not involve kinetic energy in that? Because during the movement, the speed must be non-zero, otherwise it's impossible to move an object.

Answer 1

If the velocity is constant the kinetic energy will not change. Also if we are not given the details such as velocity after work, how can we talk about the change in kinetic energy? In your problem it is given that (or you have to assume that) the object is moving with a constant velocity (I am here referring to your second problem), thus you have nothing to worry about change in the kinetic energy. However, when talking about the engine, it gives a velocity to the water as well as potential energy. As water had zero velocity before, the kinetic energy increases surely.

Answer 2

Good question. The problem is an incompletely described situation. Such questions are common, and it's up to the reader to make reasonable assumptions about the missing information.

In the case you present we are told the speed and that it's constant. There are several possibilities consistent with that wording. One is that it never starts and never stops. Another is that we are interested only in the inteval immediately after it starts and immediately before it stops. This is the usual assumption in projectile motion problems. Those questions typically ask "what is the final velocity". Well, if it hits the ground and stops the final velocity is zero. That's not what's meant. Another possibility is that we do include stopping and starting. In that case the gain in KE at the begining is lost at the end. In all of those cases the change in KE is zero.

Here are two more possibilities: 1.) it starts at zero speed but passes through the end location at 2 m/s and 2.) motion begins at 2 m/s before we start paying attention, but stops at the top. In these two cases the KE does change, and we do need to consider it. I suggest that these two possibilites are less reasonable than the three above. These two lack the symmetry that the first three have.

The reasonable assumption is KE is constant.

Answer 3

I think there are already good answers from ACB and garyp, but I would like to add in the discussion with some calculations.

In general, we know by energy-work that the change in kinetic energy is equal the work done by/over a system $$ \Delta E_{kin} = W. $$ This can be showed straight from Newton's second law and the definition of work, so quite general theorem, no matter conservative forces or not, etc. When we ask

Find the change in energy when an object moves upwards by a given height

It's tacitly assumed that the object will move upwards by itself. Since gravitation is a conservative force, we know that $$ W_{grav} = - \Delta U $$ and by the least result, $$ \Delta E_{kin} = - \Delta U \quad \Rightarrow \Delta E_{kin}+\Delta U = 0 $$ which is basically the conservation of total energy.

---

Now consider the problem

An engine draws water from depth of 10m with constant speed 2m/s at rate of 1kg/s . What is the power of the engine?

Of course the work-energy theorem apply here, so the first equation is true. But we have no reason to believe that the total energy for a volume of water is conserved. It happens since it is not moving by itself, but there is an engine here. But we can find the answer by carefully applying the energy-work theorem.

The secret is to find all the forces acting on a volume element of water, and we know there are two: the gravitational force and the external force of the engine. Both forces realizes work. Since both forces are in different directions, we have $$ \Delta E_{kin} = W_{eng} - W_{grav} $$ Here I considered as positive the work that contributes to the movement of drawing. We know how to write the gravitational work, so $$ \Delta E_{kin} + \Delta U = W_{eng} $$ And this is what you need to solved, as you already solved.