# Is work-energy theorem valid for Power = force * velocity?

Work energy theorem states "work is said to be done when there is change in Kinetic energy", i.e. $$W = ∆KE$$

but as we know, $$P=\vec F \cdot \vec v$$, where a constant force $$\vec F$$ maintains a constant velocity (overcoming opposing force). Existence of power implies existence of work.

Since the velocity is constant, $$∆KE=0$$, which means there is no work. So, what is wrong here? Please correct me.

• If the velocity is constant then the net force is zero. So, while one force may be doing positive work, the other forces that equilibrate it are, on net, doing negative work that exactly balances it. Apr 25, 2020 at 11:43
• Enter this as an answer! Apr 25, 2020 at 11:52

Work energy theorem states "work is said to be done when there is change in Kinetic energy", i.e. Work = ∆Kinetic energy

Actually, the theorem is more accurately stated as:

The net work done on an object equals its change in kinetic energy.

Notice the emphasis on the word "net". That's because work can be positive or negative and if the amount of positive work done on an object equals the amount of negative work done on the object, the net work is zero and there should be no change in kinetic energy. That is the case if the velocity of an object is constant. Per Newton's first law, either there are no forces acting on the object, or the net force of several forces is zero, meaning any positive work done by forces equals the negative work done by other forces.

An example is an automobile moving at constant velocity. The positive work done by the engine moving the car forward is countered by equal negative work done by dissipative forces (air resistance, rolling resistance of the tires, etc.) acting in the reverse direction, for a net work of zero and an change in kinetic energy of zero. In effect, all of the work done by the engine winds up dissipated as heat due to friction.

Hope this helps.

Work energy theorem states "work is said to be done when there is change in Kinetic energy", i.e. Work = ∆Kinetic energy

This is not a very accurate statement of the work energy theorem. A more correct statement is that the “net work” is equal to the change in kinetic energy: $$W_{net}=\Delta KE$$

This is an important distinction because the term “net work” has a specific and very restricted meaning: $$W_{net}=\vec F_{net} \cdot \vec{\Delta x}_{COM}$$ where $$\vec{\Delta x}_{COM}$$ is the displacement of the center of mass.

Now, it is perfectly legitimate to take the time derivative of this to get $$P_{net}=\vec F_{net} \cdot \vec v_{COM}=\frac{d}{dt}KE$$. This would be the power version of the work energy theorem.

Power= force* velocity, where a constant force F maintains a constant velocity (overcoming opposing force). Existence of power implies existence of work.

Here you are mixing up two different powers. The first is the “net power” from the work energy theorem. Since there is an opposing force the net force is zero so $$P_{net}=\vec 0 \cdot \vec v = 0=\frac{d}{dt}KE$$. So indeed the KE is constant and the object moves at a steady speed.

The second power is the power from an individual force. Sometimes this is called mechanical power or thermodynamic work, depending on the context and the preference of the author. This can be written $$P=\vec F \cdot \vec v$$ where $$\vec F$$ is the individual force and $$\vec v$$ is not the velocity of the center of mass but rather the velocity of the material at the point of application of the individual force.

In your case the individual force has a non-zero power even while the “net power” is zero. The difference between the two implies that either the opposing force has an individual power that is equal and opposite the first force’s power or the internal energy of the object is changing.