Work - constant or zero speed

Question

Say I'm riding a bicycle at speed $v$. There's air that causes drag force, let's suppose it's constant, equal $F$, and it doesn't change with speed (we know it does increase quadratically with speed, but it doesn't matter here).

Now, is there any work done by air drag force? The formula says force times distance. We have distance here, because the bike is moving, and force as well. I need to do more work to maintain my speed because of the air drag force, so I'm certainly adding some energy to the bicycle, and the air drag is removing that energy at the same time (thus the speed is constant). The net work is zero, but when looking from the perspective of individual forces, there IS identical non-zero work on both sides, but with opposite signs.

Now, let's consider a different situation: me and my friend are pushing a box, standing on opposite ends of the box with force $F$. The box isn't moving, but is there any work done by me or my friend? You could say - there's no work, because all the energy your muscles generate is transfered into heat.

But what if the box WAS moving, say, at constant speed $1$ kph? Would there be work then, even though the net force applied to the box by me and my friend was $0$?

There are many places on the Internet where you can find people arguing the can be non-zero work (looking from the perspective of a single force) even if the net force is $0$ (and net work is $0$ too), like these:

https://www.physicsforums.com/threads/work-done-on-object-when-net-force-is-zero.619110/

Answer 1

Let's say that you are standing on the ground and you observe two horizontal forces of equal magnitude, but acting in opposite directions on the block, because of which the net vector sum of all horizontal forces is zero, i.e. $$\sum F_x=0 \implies \left(a_{x}\right)_{net}=0 $$ Thus the block moves with constant velocity. Now, you wan't to calculate the net work done by the forces. The equation for work done by force is given by, \begin{align} W =&\int\limits_{x=x_1}^{x=x_2}\vec{F(x)}\cdot \vec{dx}\\ &=\int\limits_{x=x_1}^{x=x_2}|\vec{F(x)}|\times|\vec{dx}|\times \cos\theta \end{align} where $\theta$ is the smaller angle between force vector $\left(\vec{F(x)}\right)$ and displacement vector $\left(\vec{dx}\right)$, when joined Head-to-Head Or Tail-to-Tail.

Now, since in your case the force vector remains constant throughout the motion, our work equation becomes, \begin{align} W&=\int\limits_{x=x_1}^{x=x_2}|\vec{F(x)}|\times|\vec{dx}|\times \cos\theta \\ & = |F|\times\cos\theta \int\limits_{x=x_1}^{x=x_2}|\vec{dx}|\\ & = |F|\times\cos\theta\times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ \end{align} One the forces, would be acting along the direction of displacement, the angle made by this force with the displacement vector would be $0^°$, let us call this force say $F_1$. The other force would make an angle of $180^°$ with displacement vector, let us call this force say $F_2$. Over here, $|F_1|=|F_2|$( magnitude of $F_1$ and $F_2$ are equal).

Other than these two forces there can be two additional forces, if we assume that the experiment is being conducted in space where there is gravity, thus a gravitational force; and a normal reaction force if the block is moving on a surface, thereby compressing it due to its weight(because of gravity).The gravitational force always act in vertically downward direction, and the normal reaction force always act normal to the surface. Over here we assume that the block is moving on a horizontal surface, thus the gravitational force and the normal reaction would be anti- parallel to each other and both would make an angle of $90^°$ with the horizontal displacement.

Since there is no acceleration in vertical direction, i.e. $$\left(\vec{a_{vertical}}\right)_{net} = 0$$ $$|\text{gravitational force}| = |\text{normal reaction force}|=|mg|$$

And since there was no initial velocity in vertical direction and the acceleration is also taken $0$, there will be no displacement in vertical direction.

Thus, work done by $F_1$ would be, \begin{align} W_1& = |F_1|\times\cos 0 \times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ & = |F_1|\times 1 \times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ & = |F_1|\times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ \end{align}

Similarly work done by force $F_2$ would be, \begin{align} W_2& = |F_2|\times\cos 180^° \times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ & = |F_2|\times -1 \times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ & = -|F_2|\times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ \end{align}

For Gravitational force, work done would be, \begin{align} W_G& = |mg|\times\cos 90^° \times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ & = |mg|\times 0 \times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ & = 0\\ \end{align}

For Normal reaction force, work done would be, \begin{align} W_N& = |\text{normal reaction}|\times\cos 90^° \times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ & = |mg|\times 0 \times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ & = 0\\ \end{align}

Therefore, now the net work done by all the forces would be sum of work done by all individual forces.

That is, \begin{align} W_{net}&= W_1 + W_2 + W_G + W_N\\ &=|F_1|\times\left(|\vec{x_2}|-|\vec{x_1}|\right) + \left(-|F_2|\times\left(|\vec{x_2}|-|\vec{x_1}|\right)\right) + 0 + 0\\ &=|F_1|\times\left(|\vec{x_2}|-|\vec{x_1}|\right) - |F_1|\times\left(|\vec{x_2}|-|\vec{x_1}|\right)\\ &= 0\\ \end{align}

Thus, $$W_{net}= 0$$

The net work done on the block by all the forces is $0$, work done by individual forces may or may not be $0$.

This is also evident from the fact that the Kinetic Energy, (given by $\frac{1}{2} m v^2$) remains constant through out the motion as the velocity of the block remains constant, because $\left(a_{x}\right)_{net}=0$ because two equal anti-parallel forces are applied on the block.

And kinetic energy being constant $\Delta K= 0$, and we know that $W_{net} = \Delta K$, thus $W_{net} = 0$

We had earlier assumed that the observer was standing stationary on the ground, now assume there is another observer who is moving with same velocity as the block. For him, in his inertial frame of reference the block hasn't moved, because the displacement of the frame and the block would be equal as they move with the same velocity.

Thus this observer will say that the work done by individual forces is $0$, thus $W_{net}= 0$.

As I had earlier said that the work done by "individual forces may or may not be $0$" it depends on the frame of reference. We will always calculate the work of any individual force in any reference frame from our basic work equation that is,

$$W =\int\limits_{x=x_1}^{x=x_2}\vec{F(x)}\cdot \vec{dx}$$

But all the observers in their respective inertial frame of reference would always agree on $\bf{W_{net}}$. That is why nature as a whole is a conserved system.

In your experiment where you are riding the cycle, you said that the cycle was moving with constant velocity $\implies \Delta K= 0$, which means that $W_{net}$ must equal to $0$. Now, if we take the rider and the cycle as our system, we observe that there are two external forces acting on the system: Air Drag and Muscle Force (excluding the gravitational force and the normal reaction from ground), we also exclude the internal forces as they work in pair according to Newton's Third Law of Motion, thus canceling out the work of each other as a whole (as we observed in the above example how $F_1$ and $F_2$ cancel out each other's work as a whole).

We can write $W_{net}$ as $W_{external} + W_{internal}$, in this case $W_{internal}=0$, as stated above. Now external forces are : Air Drag and Muscle Force.

Caution: Don't confuse Muscle force as an Internal force, we regard it as an internal force, because any organism (like human being) can generate energy whenever it wants from the fat and other nutrients inside body, thus it is regarded as an external force. You can look at it as $\text{Sun's light energy }\underrightarrow{photosynthesis }\text{ food }\underrightarrow{ digestion }\text{ ATP = Muscle Energy}$ thus in fact Muscle energy is Sun's Energy, an external source.

now, \begin{align} & W_{net} = W_{external} = W_{Air Drag} + W_{Muscle} = \Delta K = 0\\ & \implies W_{Air Drag} + W_{Muscle} = 0 \\ & \implies W_{Air Drag} = - W_{Muscle}\\ \end{align}

Thus the rider is trying to keep the energy of the system constant(for a constant velocity) by continuously compensating the energy lost due to Air Drag by putting in his muscle energy.

Answer 2

If you do not know calculus or vectors, I have a review of both that I can insert here in this answer. I firmly believe that trying to understand Newtonian mechanics without some understanding of calculus and vectors is a Bad Idea, in the sense that you can turn the first few weeks of such a physics course into a vectors-plus-calculus course and cover objectively more material overall.

Power, work, and why we care.

The power flowing into or out of a particle is defined as the dot product between its velocity and the sum of the forces upon it, $$P = \vec v \cdot \sum_i \vec F_i.$$ Here the sum-of operator $\sum_i$ just means, "I want to sum this over all numbers $i$, hence all forces $\vec F_i$, that are acting on the particle." The dot product here means that forces perpendicular to the velocity do not, in general, "count", nor does anything if the velocity is zero.

We care about this quantity because one of Newton's laws is that $\sum_i \vec F_i = m \frac{d\vec v}{dt},$ where $m$ is the mass. This means that the power can also be expressed as:$$P = m \frac{d\vec v}{dt} \cdot \vec v = \frac {d}{dt} \left({1\over 2} m \vec v \cdot \vec v\right) = \frac{dK}{dt}$$where $K$ is the so-called kinetic energy $K = \frac 12 m v^2.$

If we integrate power over time to find "work", the $\vec v ~ dt$ collapse into a parameter $d\vec\ell$ and we get what's called a "line integral" along a path $\mathcal P$,$$W = \int_{t_0}^{t_1} dt~P(t) = \int_{t_0}^{t_1} dt~\vec v\cdot \sum_i \vec F_i = \int_{\mathcal P} d\vec\ell\cdot\sum_i\vec F_i.$$Since the power is the time-derivative of the kinetic energy, and work is the time-integral of power, we have $W = K(t_1) - K(t_0)$, the so-called work-energy theorem, as a special case of the fundamental theorem of calculus.

The biggest reason that we care is that for some forces, the rules are simple enough that any path $\mathcal P$ which closes upon itself can be proven to do $0$ work: these are called "conservative forces" and we can use the work to define a "potential energy function" over space: $W = U(x_0, y_0, z_0) - U(x_1, y_1, z_1)$, defined "backwards" so that a negative work is done when you go "uphill" on the potential energy function, or a positive work is done when you go "downhill", just like how actual hills work. Then we get a principle of conservation of energy; $K_1 - K_0 = U_0 - U_1$ is the same as $K_1 + U_1 = K_0 + U_0$, so the value $K + U$ remains the same no matter what else happens.

Can we refer to the "work due to a force" when there are multiple forces?

The answer here is a qualified yes. Just as a force is a tendency to accelerate that might have to fight with other forces (other tendencies) before we find out how things actually accelerate, we can use the same "tendency" trick to think of the work-due-to-a-force as a tendency to change kinetic energy, rather than an actual change in kinetic energy.

Once you make this shift in thinking, you can talk about the power flowing from a force or the work done by a force without any problems, because defining $P_i = \vec v \cdot \vec F_i$, the "linearity over sums" property of the dot product tells us that indeed, $P = \sum_i P_i.$ The same linearity also is a property of time integrals; $\int dt [f(t) + g(t)] = \left[\int dt f(t)\right] + \left[\int dt g(t)\right],$ so we also have $W = \sum_i W_i.$ Easy peasy.

The nicest result here is that the sum effect of a bunch of conservative forces is the same as one conservative force whose potential energy is just the sum of their potential energies.

How power flows change under a change of reference frame.

At one point, a certain Mythbuster (people on a TV show on the Discovery channel who blow stuff up and also crash lots of cars) accidentally stated an incorrect fact: if you see two cars crashing into other at speed 50 mph, he said, this would have more carnage than simply one car crashing into an immovable wall. This was apparently based on the misguided intuition that the relative velocity is twice as large, which produces a kinetic energy four times as large, divide that evenly among each car, and you have that each car needs to do double the work it does (by crunching) to slow down.

Many people wrote in to tell him he was wrong, so he did an episode where he crashed some more cars, and then we saw graphical footage proving that he was wrong. Before I tell why he was wrong, you might want to step back and figure out: do you see the fallacy he was making here?

Last chance to hide this page and work it out yourself.

If you're still stuck, the basic principle is that, if you change coordinates to one where one car is stationary and the other car is moving with velocity $2\vec v$, then afterwards the two cars are travelling together at velocity $\vec v$ in these coordinates. So we take this four-times-as-much-kinetic energy, but we have to split it up: half goes into crunching, half goes into the output motion of the two cars. The crunching energy per car is therefore only a quarter of the total energy, so it's the exact same as hitting an immovable wall.

So that's a guiding principle if you're going to analyze power or work. If you change into a reference frame travelling at speed $\vec u$ relative to your old one, yes, all of the powers change; $P_i' = (\vec v - \vec u)\cdot \vec F_i$ is different from $P_i$ creating an aggregate difference $P' = P - \vec u \cdot \sum_i \vec F_i.$ While the power is basis-independent it is nonetheless affected by changes in reference frame.

But, that's not troubling to the work-energy theorem, because after all the kinetic energy changes to $\frac 12 m |\vec v - \vec u|^2 = \frac 12 m (v^2 - 2 \vec v \cdot \vec u + u^2).$ The time-derivative of this is $$\frac{dK'}{dt} = \frac{dK}{dt} - \vec u \cdot m \frac{d\vec v}{dt} = \frac{dK}{dt} - \vec u \cdot \sum_i \vec F_i.$$The kinetic-energy-change picks up exactly the same term that the power expression had, perfectly canceling each other out.

Application to your question

After all is said and done, we have discovered that work and power are not objective in the way that you want them to be, but depend on a choice of reference frame. However we have also found that the work/energy way of doing physics is objective.

In the reference frame which sees a cyclist flying past, we say that she is doing positive work to counteract the negative work due to air drag. She might instead say that in her reference frame no work is being done, but she still has to struggle against the "wind" that wants to push her backwards. These different descriptions are both valid.

You and a friend are pushing a box equally; we can say that you are not doing any work, sure. Now we locate you both on a sailboat travelling at some $\vec u$ relative to us, fine, you're both doing works which cancel out. It's no big deal.

Answer 3

1. Work is a scalar and does not have a direction.
2. If your box was moving, because of the forces you apply than your forces may be equal in value but surely not opposite in direction or spot of application or both. In this case you would count the work of a sum of two forces, which cant be zero in any case if the box is moving.
3. If the box is moving at constant speed even if the total force of you and your friend is zero, the total work of it regarding to the box is also zero, because nothing stops or accelerating the box which stays at rest.