I agree with CuriousOne that the example is more confusing than helpful, but this is the way I would explain it.

Suppose you take a spring, place it on the ground then compress it. If you now suddenly let go of the spring it will rebound and bounce upwards off the ground:

The spring clearly has work done on it because its kinetic energy increases and that increase must have come from somewhere. However the ground can't have done any work on the spring because the ground hasn't moved. It should be obvious that the potential energy in the compressed spring has been converted into kinetic energy of the uncompressed spring - in effect the spring has done work on itself. This is what your book means by:

transfers of energy from one type to another inside the object

i.e. potential energy of the compressed spring has been converted into kinetic energy of the uncompressed spring.

In the case of the skater the skater's arms correspond to the spring and the rail corresponds to the ground. The skater's arm isn't a spring, of course, because it's chemical energy not potential energy being converted to kinetic energy by the skater's muscles. Nevertheless the same principle applies.

Here's an example of why this is done using a gravitational context.

## Kinetic energy

The kinetic energy of a system of particles is *defined to be* $K_\text{tot} = \sum_i K_i$, where the sum runs over all particles in the system. This is just a definition, and choosing to use such a definition allows one to speak of the kinetic energy $K_i$ of the $i$th particle, for example, since it's just one of the terms in a sum.

Okay, that was boring.

## Potential energy

Gravitational potential energy $U_\text{g} = -\frac{Gm_im_j}{d_{i,j}}$ is defined for *pairs* of particles.$^\text{1}$ Mathematically you can see that in the subscripts of the masses $m_i$ and the distance between the two particles $d_{i,j}$. Why? Because to determine potential energy from scratch, you must consider how two objects interact, which in classical mechanics is where *forces* come in. And forces themselves come in pairs.

Now, the fact that they come in pairs means one must be careful when determining the potential energy in some given situation, otherwise you might **over-count.** Suppose, for example, there are 4 particles with masses $m_i$, and the distance between the $i$th and $j$th objects is $d_\text{i,j}$. What, then, is the potential energy of the masses and distances are all given? The answer is
$$U_\text{g}=-G\left( \frac{m_1m_2}{d_{1,2}} + \frac{m_1m_3}{d_{1,3}} + \frac{m_1m_4}{d_{1,4}} + \frac{m_2m_3}{d_{2,3}} + \frac{m_2m_4}{d_{2,4}} + \frac{m_3m_4}{d_{3,4}} \right)$$
First, note that since there are six terms you can't make a one-to-one associating of particles with potential energy. Second, if someone *tried* to associate potential energy with a single particle, then that person might mistakenly include too many terms in their sum:

$$U_\text{g}\ne-G\left( \underbrace{\frac{m_1m_2}{d_{1,2}}}_\text{a term for particle 1} + \underbrace{\frac{m_2m_1}{d_{2,1}}}_\text{a term for particle 2} + \cdots \right)$$
This incorrect expression would over-count the number of terms in $U_\text{g}$.

To avoid such over-counting, and to make our language more inline with the mathematics, we tend to speak of the potential energy of the *system.*

$^\text{1}$The expression $-\frac{Gm_1m_2}{d_{1,2}}$ is a more general form of the more-common $U=mgh$, but that's a separate issue.

## Best Answer

The

potentialis defined as a function $U$ such that the conservative force $\vec F$ that we are studying is given by the gradient $\vec F = -\nabla U.$ Since you probably have not seen vector calculus yet, let me be very careful to write this out as the components, $$F_x = -\frac{\partial U}{\partial x},\\ F_y = -\frac{\partial U}{\partial y},\\ F_z = -\frac{\partial U}{\partial z}.$$These "partial derivatives" are evaluated as normal derivatives treating the other variables as constant, so for example the potential $U=k~x^2~y + p~z^2$ would generate $F_x = -2k~x~y, F_y = -k~x^2, F_z = -2p~z.$Partial derivatives are the natural way to understand calculus on a function of many variables. In single-variable calculus, you were trying to approximate a curve with a tangent line; in this multi-variable calculus we are trying to approximate surfaces with planes. In particular, if you

canmake this approximation then it means that a function can be expanded around a point, $$f(x + \delta x, y+\delta y, z+\delta z)\approx f(x, y, z) + \frac{\partial f}{\partial x}~\delta x+ \frac{\partial f}{\partial y}~\delta y+ \frac{\partial f}{\partial z}~\delta z.$$Here by $\delta q$ I just mean "a little change in $q$", whatever $q$ is. One could also move this term $f(x,y,z)$ to the left-hand side and refer to that difference as $\delta f$, if you'd like. This understanding of expanding out a multivariable function will be important.The

powerexerted by a force on a particle is the dot product of that particle's velocity with the force, and theworkdone over the path is the time integral of power exerted by that force. It is common to denote the position of the particle as a vector $\vec r(t)$ with components $r_{x,y,z}(t)$ and then this is: $$ P(t) = F_x~\frac{dr_x}{dt} + F_y~\frac{dr_y}{dt} + F_z~\frac{dr_z}{dt}\\ W = \int_{t_0}^{t_1}dt~P(t).$$ The combination of these two definitions is what you seem to be asking about: but it is not very complicated at all. Combine the two and then stare for a second at the following: $$P(t)~\delta t = -\frac{\partial U}{\partial x} ~\frac{dr_x}{dt}~\delta t -\frac{\partial U}{\partial y} ~\frac{dr_y}{dt}~\delta t -\frac{\partial U}{\partial z} ~\frac{dr_z}{dt}~\delta t.$$What should now stand out to you is that this is very much like the above expression for $\delta f$ above, if we defined $\delta x = \frac{dr_x}{dt}~\delta t$ and so on for $\delta y, \delta z.$ And those are very natural definitions, as $r_x$ represents an $x$-component of position and if we take this time-derivative we get a component of velocity, and multiplying against a short time $\delta t$ we get a small change in this $x$-component due to the particle's current motion.There is a more formal way to do this and it is to invoke the

chain rule, which says that when we apply some function $U(x, y, z)$ to these time-varying components $x = r_x(t)$ and so forth, we find that:$$\frac{d}{dt} \Big(U\big(r_x(t),~r_y(t),~r_z(t)\big)\Big) = \frac{\partial U}{\partial x}~\frac{dr_x}{dt} + \frac{\partial U}{\partial y}~\frac{dr_y}{dt} + \frac{\partial U}{\partial z}~\frac{dr_z}{dt}.$$ Therefore what we have found above is simply, $$P(t) = - \frac{d}{dt} \Big(U\big(r_x(t),~r_y(t),~r_z(t)\big)\Big).$$The work is the time integral of power, but integrals perfectly undo derivatives, and therefore when we do this definite integral we get from the fundamental theorem of calculus, $$\begin{align} W &= -\int_{t_0}^{t_1} dt~\frac{d}{dt} \Big(U\big(r_x(t),~r_y(t),~r_z(t)\big)\Big) \\ &= -\Big(U\big(r_x(t_1),~r_y(t_1),~r_z(t_1)\big) - U\big(r_x(t_0),~r_y(t_0),~r_z(t_0)\big)\Big)\\ &=-\Delta U.\end{align}$$ That's really all there is to it: for conservative forces $\vec F = -\nabla U$ the power $\vec F \cdot \frac{d\vec r}{dt}$ is immediately seen to be a chain rule expression $\frac{d}{dt} U(\vec r) = \nabla U\cdot \frac{d\vec r}{dt}$, which identifies it as a total time derivative, and therefore the work, which is just the time integral of the power, must be the overall change in the quantity: in this case the quantity is $-U$ and so $W = \Delta(-U) = -\Delta U.$