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I came across this definition for potential energy:

If we let $T$ be the Kinetic energy, we have that:

$$T = \frac{1}{2}mv^2 \implies T = \frac{1}{2}m{x'}^2$$

$$T'= mx'x'' = F(x)x' \implies \\T = \int F(x)x'dt = \int F(x) dx \tag{2.4}$$

Then, the book says that if we define

$$V(x) = -\int_{x_0}^{x}F(x')dx'$$

we can write the equation:

$$T+V = E$$

where $x_0$ is an arbitrary constant, corresponding to the arbitrary additive constant of integration in (2.4)

I understand that potential energy can be thought as everything from total energy $E$, such that this everything is not the Kinect energy. However, I don't understand why the integral $$V(x) = -\int_{x_0}^{x}F(x')dx'$$

Works and can be understood as potential energy. Could someone explain better to me why $T+V = E$, where $E$ is the total energy?

Is there an example for this?

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  • $\begingroup$ Why are there no limits in the integral 2.4? Shouldn't these limits also be x_0 and x? But then T+V would always add up to zero. That's a bit confusing here. $\endgroup$
    – Merlin1896
    Jun 20 '15 at 22:53
  • $\begingroup$ The x' in the definition of V(x) is NOT the derivative of x. It's a dummy variable. You might as well replace it by 'y' $\endgroup$ Jun 21 '15 at 9:14
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The whole energy-concept is a reformulation of Newtons laws. Starting from $\vec{F}=m.\vec{a}$, you could wonder about the effect of a force during a displacement. You call the concept 'work' and do the math

$$W=\int_{\vec{x}_1}^{\vec{x}_2} \vec{F} d\vec{x} = \ldots = \frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2$$

To save yourself some work you define

$$T = \frac{1}{2}mv^2$$

So you can write

$$W = T_2 - T_1$$

To calculate the left side of the equation, you need to know every detail (every $\vec{F}(\vec{x})$), but (part of) the beauty of the whole energy-concept is that for the right side, you only need to know the end ($v_2$) and the beginning ($v_1$).

This leads to the idea of reshaping as much as possible from the left side of the equation, in such a way that its effect (the work done) is reduced to knowing only beginning and ending (just a matter of being efficient/lazy). This only works for conservative forces (like gravity, electric forces, ideal springs), as here the work done is only dependent on start-position and end-position (wether you fall from a tree branch 2 metres high, or slide down from a slide 2 metres high, the work done is always the same). For forces like friction, it is clear that the work done depends also on the trajectory, not only on start and finish.

So it finally becomes

\begin{align} W &=& \Delta T \\ \Rightarrow W_{cons \ forces} + W_{non\ cons \ forces} &=& \Delta T\\ \Rightarrow W_{ncf} &=& \Delta T - W_{cf}\\ &=& \Delta T + \Delta V\\ \end{align} where $$\Delta V = -W_{cf} = -\int_{x_1}^{x_2}F_{cf} dx$$ So potential energy is just a reshaping of the work done by a conserative force to gain efficiency (and often insight, as it provides a slightly different angle of looking at things).

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  • $\begingroup$ $\Delta V = -W$? Shouldn't it be $W_{ncf}$? $\endgroup$
    – guerlando
    Jun 21 '15 at 3:28
  • $\begingroup$ Compare the 2 last equations: $W_{ncf} = \Delta T - W_{cf}$ and $W_{ncf} = \Delta T + \Delta V$. These are the same, but with different 'names'. For these equations to be the same, it should be that $\Delta V = -W_{cf}$. $\endgroup$
    – Dries
    Jun 21 '15 at 10:09
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Your presentation is a little confused. I would read your book again very carefully, taking care to understand the meaning of each variable.

The integral that you write for potential energy is the definition of potential energy, if $F$ is taken to be an internal force, that is, a force between two objects within the system. For example for free fall at the surface of the earth the system can be taken as two objects: a ball and the earth, with $F$ being the gravitational force.

In your first paragraphs, the meaning of $F$ is different. In that case, $F$ is an external force; a force due to an agent outside the system. Again using the free fall example, the system might be the single object, the ball, with gravity then being and external force.

How the math is set up depends on what you choose to be your system. Is the system the earth/ball pair? or is it a single object, that ball? The solution to the dynamics will be the same, but the "bookkeeping", or how the equations are set up, will be different. If you take the system to be the two objects earth and ball, then gravity is an internal force, and one needs to use potential energy. If you take the system to be the ball with gravity an external force, then potential energy cannot be defined at all!

Be sure you understand the meaning of each symbol, and be sure you keep in mind what you have chosen to be your system.

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