1
$\begingroup$

This is how my book explains it:

$W_{net} = \delta K$ Since the net work is tied to changes in kinetic energy and changes in speed, a mass must accelerate in order for net work to be nonzero. Thus, when an object moves at constant velocity, the net work is always equal to zero.

If "net work" is understood to be the amount of work done, or just whatever is left after things are cancelled out, I don't think then there has to be any acceleration of increase in speed for there to be net work. Like when I lift something up to increase its potential energy, at last the object remains stationary. But there is net work in form of gains in potential energy.

$\endgroup$

4 Answers 4

2
$\begingroup$

The total work done by all forces acting on an object throughout the motion interval of interest is what the work-energy principle involves. It never says "no forces are doing work." And it doesn't talk about changes in potential energy. The changes in potential energy are involved in the work done by the conservative force attached to the particular potential energy. The bookkeeping of work, (force and direction and distance), can become tedious in some situations, but it works.

Let's say a person lifts a stationary box from a floor, carries it somewhere in the same room and sets it on a table. Let's also ignore air resistance, etc. The initial KE of the box in the table/room/floor reference frame is zero. The final KE of the box is zero. The total work done by all forces acting on the box is zero.

What forces did work on the box? AHA! The normal and frictional forces of the person's hands initially did positive work (force is up, motion is up), then more positive work as the person exerted a sideways force with sideways motion, and finally some negative work as the person exerted upward force to gently place the box on the table rather than dropping it. Gravity initially did negative work (gravity down, box moves up), and zero work as the box moved sideways (ignore the slight bouncing of the box while it's being carried) then positive work while the box moved down. All this work on the box adds to zero.

Meanwhile, the heart and brain and muscles were all doing work internal to the person, so they had to go get a soda pop and sit down and rest.

$\endgroup$
1
  • $\begingroup$ What about vaporizing water? There is no increase in kinetic energy but only in potential energy. What is the opposing force in this case? $\endgroup$ Aug 4, 2015 at 18:21
1
$\begingroup$

When you lift an object (with no increase in kinetic energy), you're doing positive work to lift it up, while gravity is doing negative work (trying to pull it down). In total, the net work is zero.

Another way to think about it: total work is net force times displacement. If the object isn't accelerating, you know from $F=ma$ that net force is zero.

$\endgroup$
2
  • $\begingroup$ But the object is accelerating? It's only stops at the end. $\endgroup$ Jul 24, 2015 at 20:24
  • $\begingroup$ It's accelerating slightly. If you want to think about it carefully, you can say it accelerates a LITTLE at the beginning, and deaccelerates a LITTLE at the end. The total work done at the beginning is a LITTLE bit positive, the work done at the end is a LITTLE bit negative. We usually just approximate this as saying there is no net force throughout. $\endgroup$ Jul 26, 2015 at 21:28
1
$\begingroup$

Net work is basically defined as $\displaystyle\int F(s) \cdot \mathbf{d}s$, where $F(s)$ is the net force – the sum of all forces acting on the body.

For example, when a car holds a constant velocity on a highway the net work done is zero (since the net force is zero) but that doesn't mean work isn't done. Both the engine and drag due to air resistance are doing work on the car, in equal amounts. However, the energy released in the engine does not change the kinetic energy of the car. It's instead converted into other forms of energy (mostly heat).

Since the kinetic energy of the car does not actually change, the work done on it must therefore be zero. The energy provided by the fuel is instead converted into other forms of energy, and does not do any work on the car itself.

$\endgroup$
1
$\begingroup$

If you accidentally drop that $1000000 rare penny onto the ground, while digging for stuff in your pocket and it makes a clattering sound for about ¼ second after hitting the ground, but you can't see it anywhere in all the clutter on the ground, then where do you look?

Assuming it fell from about a meter on high, then its gravitational potential per unit mass was close to 10, in metric units. When it hit the ground, the potential energy was converted into kinetic energy, the square of its speed was 20, in metric units, and its speed was about 4½ meters per second.

Assuming that it didn't keep rolling quietly, after it stopped making that clattering sound, then it came to a stop in ¼ second. The sound was the energy of its motion getting lost to the air, so at no time did it ever move faster than the speed it had when it hit the ground. Therefore, you may search for the coin within a 1⅛ meter radius of the spot where you heard it hit the ground and (assuming it didn't quietly roll off) will find it in there, somewhere.

The greatest likelihood is that it will be at about ⅔ that distance (i.e. ¾ meter) or less from the point where you heard it hit the ground, since - roughly speaking - it will have lost its kinetic energy at a constant rate.

That's pretty useful, as an application. Therefore, the work-energy theorem is useful.

$\endgroup$

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