How can we show Power = $\mathbf{F}\cdot \mathbf{v}$?

Question

How can we say that $$\text{Power} = \mathbf{F}\cdot \mathbf{v}$$

We know that small work done by a force $\mathbf{F}$ to displace an object by '$\mathbf{x}$' is

$$W = \mathbf{F}\cdot \mathbf{x}$$

So derivating wrt time, we get

$$\begin{align} P=\dfrac{dW}{dt}&=\frac{d\mathbf{F}}{dt}\cdot \mathbf{x}+\mathbf{F}\cdot\dfrac{d\mathbf{x}}{dt}\\ &=\frac{d\mathbf{F}}{dt}\cdot \mathbf{x}+\mathbf{F}\cdot\mathbf{v} \end{align}$$

We get this wrong result. How actually can we show $P=\mathbf{F}\cdot\mathbf{v}$ ?

Edit

Actually I know that total work $W$ is $\int \mathbf{F}\cdot d\mathbf{x}$.

Infinitesimal work done by $\mathbf{F}$ to displace body by $d\mathbf{x}$ will be $dW = \mathbf{F}\cdot d\mathbf{x}$, so dividing by $dt$ on both sides gives $$P =\dfrac{dW}{dt} = \mathbf{F}\cdot \frac{d\mathbf{x}}{dt}$$

But I wanted a proper proof not involving differentials!

Why do you want/need a proof without 'differentials'; your own derivation uses 'differentials'? Work is literally defined as $W=\int F dx$. — nluigi, Jul 18 '17 at 10:51
Hi samjoe, the answer by Luke to the first linked duplicate "What does the equation P=F⋅v really mean?" is what you are looking for. In brief, the forces on a particle are $F_i$ and their "partial powers" on its motion are defined as $P_i=F_i\cdot v$, then we find $P=\sum_i P_i = m a\cdot v=m\frac{dv}{dt}\cdot v=\frac{d}{dt}\left(\frac12 m v^2\right),$ where the last equality is easiest to prove in the other direction (i.e. just take the derivative with the product rule and confirm that they are equal). — CR Drost, Jul 18 '17 at 14:30
Then we see that $P=dK/dt$ and so we want to integrate $W=\int P~dt=\Delta\left(\frac12 m v^2\right),$ defining the work done. Well, we derive that this integral is just $W=\sum_i \int F_i\cdot dx$ from the $dx=v~dt$ property, and so there are "partial works" due to each of the forces acting on the trajectory which sum up to a "total work", too, and that work tells us the difference in kinetic energy between the beginning and ending of that trajectory. — CR Drost, Jul 18 '17 at 14:33

score 4 · Accepted Answer · edited Jul 18 '17 at 14:05

4

The work $W$ is not equal to $F\cdot x$ generally. The correct form is $$W=\int F \cdot dx$$ So $dW=F\cdot dx$ and $P=\dfrac{dW}{dt}=F\cdot\dfrac{dx}{dt}=F\cdot v$

edited Jul 18 '17 at 14:05

jonsno

319

answered Jul 18 '17 at 10:01

endemez

56

Hm Thanks but how can we show this mathematically? I mean not by dividing with differentials? – jonsno Jul 18 '17 at 10:02
2

The answer is already mathematical. The definiton for the work you use is only valid when the force is constant. – endemez Jul 18 '17 at 10:06
@samjoe : The definition of Work done that you have written is not "mathematically" correct. It holds for 1 dimension only. – Icchyamoy Jul 18 '17 at 10:35
@samjoe that's a good enough answer. – Wrichik Basu Jul 18 '17 at 10:36

score 2 · Answer 2 · edited Jul 18 '17 at 11:02

2

See that work is done only when there is a displacement. So if there is no displacement then even having a variable force will not account to any Power. Hence the first term is excluded.

$$ W = \int \mathbf F \cdot {\mathbf v} dt $$ $$P=\dot{W} $$

When applied to the integral along with fundamental theorem of calculus it gives

$$P=\mathbf{F\cdot v}$$

edited Jul 18 '17 at 11:02

Yashas

7,163

answered Jul 18 '17 at 10:14

Icchyamoy

422

1

Yes I understand that, but how to show this without using differentials? – jonsno Jul 18 '17 at 10:39
@samjoe : I tried to write it mathematically. See if it helps or not. – Icchyamoy Jul 18 '17 at 10:48
@samjoe : see the answer of joshphysics who wrote it down explicitly. – Icchyamoy Jul 18 '17 at 11:00
Thanks for your answer and that link! I now understand completely! – jonsno Jul 18 '17 at 13:44

How can we show Power = $\mathbf{F}\cdot \mathbf{v}$?

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