# Energy and force, the basic formula

Energy and force, the basic formula

Can someone explain to me this chain of equalities?

Namely the $$v$$ in the 4th term and the next term: $$\int d/dt \ ((m/2) v^2) \ dt$$

$$E = \int F \ ds = \int m a \ ds = \int m (\dot v) v \ dt = \int d/dt ((m/2) v^2) \ dt = (m/2) \ v_0^2$$

• It is really mostly just an observation that $$\frac{\mathrm d\ }{\mathrm dt}\left(\frac12mv^2\right)=m\dot vv$$ just expressed in reverse order Commented Aug 13 at 8:26
• It's just written down in a terribly clumsy way, as if the author did not understand it him- or herself. I hope this is not from a textbook? Commented Aug 13 at 9:21

The chain of equalities that you ask about is haphazard. It's the type of arduous sequence that you get when someone is trying to get to a known end result, without knowing how to get there efficiently.

Here is how it should be done:

In preparation:

$$v = \frac{ds}{dt} \ \Leftrightarrow \ ds = v \ dt \tag{1}$$

$$a = \frac{dv}{dt} \ \Leftrightarrow \ dv = a \ dt \tag{2}$$

Going from (3) to (6):
First (1) is used to change the differential from $$ds$$ to $$dt$$, with corresponding change of limits. Next (2) is used - with change of limits - to arrive at (6).

$$\int_{s_0}^s a \ ds \tag{3}$$ $$\int_{t_0}^t a \ v \ dt \tag{4}$$ $$\int_{t_0}^t v \ a \ dt \tag{5}$$ $$\int_{v_0}^v v \ dv \tag{6}$$

It follows:

$$\int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \tag{7}$$

With the above prepared: take $$F=ma$$, and on both sides integrate with respect to position coordinate:

$$\int_{s_0}^s F \ ds = \int_{s_0}^s m \ a \ ds \tag{8}$$

Use (7) to process the right hand side:

$$\int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \tag{9}$$

The sequence of steps above highlights an underlying symmetry; $$v=ds/dt$$ and $$a=dv/dt$$ are the same pattern.

The reason that the integral $$\int_{s_0}^s a \ ds$$ can be processed is that acceleration $$a$$ as a function of time and position $$s$$ as a function of time are connected by differentiation.

The integral should be processed as an integral with a starting point and an end point, because Energy is defined in terms of difference of Energy. The choice of zero point of energy is an arbitrary choice. In any calculation in which energy is used it is difference of energy between two states that counts, not the value itself.

We know that $$a=\dot v=\frac{dv}{dt}$$ and $$v=\dot s=\frac{ds}{dt}$$.

Thus we can substitute $$a=\dot v$$ from the first identity and $$ds=vdt$$ from the second one.

If you differentiate $$\frac{1}{2}mv^2$$ with respect to the time $$t$$, you will get, using the chain rule for derivatives, that

$$\frac{d}{dt} \left(\frac{1}{2}mv^2\right) = mv\frac{dv}{dt}=mv\dot v$$

But this is exactly what we had as integrand, so we know how to evaluate the integral.

Notice that

$$\int d(???)=???$$

$$E=\int m \,\dot v\, v \ dt = \int \frac{d}{dt}\left(\frac m2 v^2\right)\,dt= \int d \left(\frac m2 v^2\right)=\frac m2 v^2\bigg|_0^{v_0}=\frac m2 v_0^2$$
$$\frac{d}{dt}\left(\frac m2 v^2\right)=\frac{d}{dv}\left(\frac m2 v^2\right)\frac{dv}{dt}=m\,v\,\dot v$$
$$E=\int\,F(s)\,ds=\int\,F(s(t))\,\underbrace{\frac{ds}{dt}}_{v}\,dt$$
and with $$~F=m\,a=m\,\dot v$$
$$E=\int m\,\dot v\,v\,dt$$