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In calculating power using the formula $\underline{F}\cdot\underline{v}$, what is the correct velocity to use? Does one use the velocity of the body on which the force is acting, or the velocity of the body providing the force? I always thought it was the former (at least because in the case of force fields the field doesn't have a velocity, so the only velocity is that of the body the force is acting on).

However, when I use this understanding on an example problem I seem to end up with results about power calculations in different reference frames that I am struggling to make sense of. I have posted this question here for you to see the numbers.

Any clarity people can provide on this point (either in general or in specific relation to the example question I posted) would be much appreciated.

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  • $\begingroup$ It might help if you would add a specific example of how you get these different results $\endgroup$
    – Hilmar
    Commented Nov 14, 2022 at 19:53
  • $\begingroup$ force times the velocity of the point where the action is applied $\endgroup$
    – basics
    Commented Nov 14, 2022 at 20:03
  • $\begingroup$ I gave a link to an example. @basics it sounds like you're confirm that the velocity of the body that the force is acting on is the correct principle. But with that principle I conclude in the example I gave that energy is flowing into a cyclist in the presence of a cross-wind, which is unsettling me. It feels a bit handwavy to say it is dissipated as heat from the cyclist, especially when without the crosswind I conclude no energy flows into cyclist, just flows through cyclist from ground to air in road frame (and vice versa in cyclist frame). $\endgroup$ Commented Nov 14, 2022 at 22:01
  • $\begingroup$ it's not a single lumped force acting on the cyclist. You should integrate the stress vector times the velocity over the whole surface of the cyclist. Assuming that the points on the surface of the cyclist move approximately at the same velocity, the result is total force times the velocity of the cyclist (times means dot product). if needed, I'll give you an answer tomorrow $\endgroup$
    – basics
    Commented Nov 14, 2022 at 22:05

4 Answers 4

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The bicycle cannot be suppling only a single force against the air. If it did it would accelerate in the opposite direction. Instead it must also create some forces against the ground. In a reference frame with the ground at rest, there is no displacement, so the energy/power from that force can be ignored. In a reference frame where the ground is not at rest, there will be power associated with that force that must be considered.

In a frame where the ground is advancing, the force from the ground onto the cyclist is positive, so there is extra energy (from the ground) that can be placed into the wind.

What is unsettling me is in one frame energy is either flowing out the ground into the air or out of the air and ground into the cyclist, but in another frame the energy is flowing out of the air into the ground or out of both into the cyclist.

That's expected. In the frame where the ground is moving forward (same direction as the relative motion of the cyclist), the cyclist is pulling energy from the motion of the ground. In the frame where the ground is moving backward, the cyclist is putting energy into the ground.

Is it right to conclude energy is accumulating in the cyclist in when there's a crosswind like in the example?

No. While the cyclist can convert some chemical energy into forces/kinetic energy, that does not happen in the reverse. Any energy from ground can only go to increase the speed of the bicycle, or be delivered elsewhere (air resistance/drag).

If the cyclist is at a constant speed in some inertial frame, then the total amount of power onto the cycles from all forces (pedals, ground, air) must be zero.

In your other link, I don't understand how X->Y and and Y->X can be anything other than the negative of each other. If the ground is doing work on the cyclist, then the cyclist is doing negative work on the ground.

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  • $\begingroup$ Thanks for your answer. You will see in the link that I gave I account for the ground and treat it as another "body" in the problem. What is unsettling me is in one frame energy is either flowing out the ground into the air or out of the air and ground into the cyclist, but in another frame the energy is flowing out of the air into the ground or out of both into the cyclist. Is it right to conclude energy is accumulating in the cyclist in when there's a crosswind like in the example? $\endgroup$ Commented Nov 14, 2022 at 21:58
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Does one use the velocity of the body on which the force is acting, or the velocity of the body providing the force?

The velocity of the body on which the force is acting. More specifically, the mechanical power delivered to a system is $\vec F \cdot \vec v$ where $\vec v$ is the velocity of the material of the system at the point of application of the force $\vec F$ on the body.

If the velocity of the material of the system at the point of application of the force is the same as the velocity of the environment at the point of application of the force, then all of the mechanical energy that leaves the environment enters the system. If the velocities differ then mechanical energy is being destroyed at the point of contact, e.g. it is converted to heat with sliding friction.

I seem to end up with results about power calculations in different reference frames that I am struggling to make sense of

Mechanical power is frame dependent, but energy is conserved in all frames. Things like the mechanical power converted to heat from sliding friction is the same in all frames, even though the amount of mechanical power transferred varies.

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Given a reference frame, the total power of a force field acting on a system is the sum of the dot product of each force and the velocity of the point where the force is acting

$\displaystyle P = \sum_i \mathbf{F}_i \cdot \mathbf{v}_i$,

being $\mathbf{F}_i$ lumped forces. If you deal with continuous distribution of forces per unit volume $\mathbf{f}$ in volume $V$, forces per unit stress $\mathbf{t_n}$ on surface $S$ or force per unit-length $\boldsymbol{\gamma}$ on line path $\Gamma$, summation is replaced by integration over the corresponding domains,

$\displaystyle P = \sum_i \mathbf{F}_i \cdot \mathbf{v}_i + \int_V \mathbf{f} \cdot \mathbf{v} + \int_S \mathbf{t_n} \cdot \mathbf{v} + \int_{\Gamma} \boldsymbol{\gamma} \cdot \mathbf{v} $.

Power of forces on a rigid body performing translation, not rotation. In this situation, the velocity of all the points of the body is constant, $\mathbf{v}(\mathbf{r}) = \overline{\mathbf{v}}$, and thus the power becomes

$\displaystyle P^{transl} = \left[ \sum_i \mathbf{F}_i + \int_V \mathbf{f} + \int_S \mathbf{t_n} + \int_{\Gamma} \boldsymbol{\gamma} \right] \cdot \overline{\mathbf{v}} = \left[ \mathbf{F}^{tot,lump} + \mathbf{F}^{tot,V} + \mathbf{F}^{tot,S} + \mathbf{F}^{tot,\Gamma} \right] \cdot \overline{\mathbf{v}} = \mathbf{F}^{tot} \cdot \overline{\mathbf{v}}$.

Kinetic energy theorem. The kinetic energy theorem states that the time derivative of the kinetic energy of a closed system equals the total power of the forces

$\dot{K} = P^{tot}$,

see https://physics.stackexchange.com/q/735204.

Kinetic energy theorem and change of reference frames. When you change the reference frame used to evaluate the position and the velocity, both the kinetic energy and the power of forces change, but the kinetic energy theorem still holds.

See https://physics.stackexchange.com/q/734777

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2D case the components of the force vector are

$$\vec F=\begin{bmatrix} F_x \\ F_y \\ \end{bmatrix}$$ the components of the velocity vector are

$$\vec v=\begin{bmatrix} v_x \\ v_y \\ \end{bmatrix}$$

where the components can be taken in any coordinate system but the same system for the force and the velocity vectors

from here

$$\vec F\cdot\vec v=F_x\,v_x+F_y\,v_y$$

analog 3D case

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