5
$\begingroup$

I don't know much yet about energy and work, as I've only had one lesson on it so far. However, our teacher told us that zero work is confusing and that we wouldn't need to know about it (with the virus and all, most units have been cut short so unnecessary things are glossed over).

Nonetheless, I still want to understand what zero work is. Which brings me to my question: From my understanding, work is equal to the force net of an object multiplied by the distance covered. In terminal velocity, the net force of an object would be 0 because the air resistance or drag of the object cancels out the force of gravity. Therefore, the product of force net and the distance covered of an object in terminal velocity would be zero. Which is why my understanding is that terminal velocity would be an example of zero work.

Apparently, my teacher isn't so sure that it is, so that brought me here. Can anyone clear up this misunderstanding, or confirm my idea?

$\endgroup$
3
  • $\begingroup$ If the net force of an object would be 0 because the air resistance or drag of the object cancelled out the force of gravity, why would your object continue to fall, rather than simply float? $\endgroup$ Jun 8, 2021 at 1:03
  • 1
    $\begingroup$ It's been a really long time since I took any simple mechanics, but since the object falling is not accelerating, it is essentially floating - if you consider your frame of reference to be something that has the same velocity as the falling object. No work is being done. $\endgroup$
    – Flydog57
    Jun 8, 2021 at 2:13
  • $\begingroup$ @RobbieGoodwin zero net force means zero acceleration i.e. no change in speed. That's why it's called terminal velocity; it's when you stop falling faster and stay at the same speed. $\endgroup$
    – Rick
    Jun 8, 2021 at 12:57

4 Answers 4

6
$\begingroup$

Gravity is doing work on you when you fall, even at terminal velocity. But, as you also point out, air drag does work as well. They are equal but with opposite signs:

  • work by gravity is positive (trying to add to your kinetic energy), whereas
  • work from air drag is negative (absorbing and removing energy from you, losing it as heat and movement of air molecules).

They cancel each other out so that the net work done on your falling body is zero.

A true no-work scenario, where no work is exchanged at all, might simply be a scenario of no velocity at all. Like, just lying still in bed. Although I must admit that I am not sure of what you are seeking as an "example of no work", since work belongs to a force and not to a scenario as such.

$\endgroup$
2
  • 3
    $\begingroup$ I have never liked the concept of "net work" because it confuses novice physics students. $\endgroup$ Jun 7, 2021 at 17:04
  • $\begingroup$ @DavidWhite Agreed. Nevertheless, work is a property associated with a single force. $\endgroup$
    – Steeven
    Jun 8, 2021 at 7:30
3
$\begingroup$

You are correct: zero total work is done on the object at terminal velocity. You could also ask "how much work is done on the falling object by gravity" and that would be nonzero. Likewise, you could ask "how much work is done on the object by the force of air resistance" and that would be nonzero. But, the total work done on the object is zero since the total net force is zero. Hope that helps!

$\endgroup$
3
$\begingroup$

When talking about work, you have to specify the force that is doing the work: forces do work. The general expression for the work $W$ done by a force $\mathbf F$ over some path $C$ is defined as an integral: $$W=\int_C\mathbf F\cdot\text d\mathbf x$$ However, in the case of a force with constant magnitude $F$ that is tangent to the path at all times, we can simplify this to just be $W=Fx$, where $x$ is the length of the path.

Now, you can choose to look at the work done by individual forces, or you can add these together to determine the net work being done on the object. It turns out that the net work done on an object is equal to its change in kinetic energy: $W_\text{net}=\Delta K=\frac12m\left(v_2^2-v_1^2\right)$, where $v_1$ and $v_2$ are the speed of the object at the beginning and end of the path respectively.

So, in your air resistance example, we can look at the work done by gravity, the work done by air resistance, or the net work. Since at terminal velocity the object's speed is constant, we can safely say that $\Delta K = W_\text{net}=W_\text{grav}+W_\text{air}=0$, and therefore $W_\text{grav}=-W_\text{air}$. The work done by gravity exactly cancels the work done by gravity.

$\endgroup$
2
$\begingroup$

Work is defined as $$W=\int\vec{F}\cdot d\vec{r},$$ which is a line integral, so it depends on the path in general. However, if the force $\vec{F}$ is conservative $(\vec{F}=-\vec\nabla V)$, the work will only depend on the initial and final points $W=V(\vec r_0)-V(\vec r_f)$. Thus, there are several situations that make the work be zero:

  1. The net force is zero $\vec{F}=0$.
  2. The force is perpendicular to the displacement $\vec{F}\perp d\vec{r}\Rightarrow\vec{F}\cdot d\vec{r}=0$.
  3. If the force is conservative $(\vec{F}=-\vec\nabla V)$ and the final and initial points are the same $\vec{r}_0=\vec r_f\Rightarrow W=0$.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.