Understanding relationship between work and energy

Question

I've read over 10 books about work and energy, and I just simply can't understand it.

First of all, they go ahead and randomly define that work is force times distance: $$W=F X \cos\theta$$ Okay, cool, this is just a definition.

Next, they go ahead and tell us that energy is the "ability to do work", then they tell us about the kinetic energy, potential energy and spring energy, all of them derived from the definition they have introduced to us ($F X$ or $\int F\,dx$). Then they tell us that the energy is conserved.

What I just don't simply understand, why is this 'ability to do work thingy' (which is energy) conserved?

They came up with a random definition and derived all the energy equations from it and then they tell us this quantity is conserved, why?

For example, why didn't they define work as $W=F X^2$ or $W=(F 2) X^3$ or anything like that, and then derive all the energy equations from there? Why is it $W=F X$? And did they come up with this conclusion from Newton's laws?

Answer 1

Sometimes when you're stuck on things, it's helpful to look at the mathematics of what's being asserted. For example, nowhere in Newton's three laws does "energy is conserved" appear.

Energy conservation does appear, however, when you have a system that behaves like $m \ddot{x}=-\nabla U$, for some function $U$, where $x$ is a position vector as a function of time. In this case it's a mathematical theorem that $\frac{d}{dt}\left(\frac{1}{2} m \|\dot{x}\|^2+U\right)=0$.

Though it's easy to get carried away and start talking about nature and systems and why some forces can be represented as $\nabla U$, in every regular mechanics book* I've read, this is what things boil down to.

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*regular mechanics as opposed to higher mechanics. In higher mechanics one states that the action $A[u]=\int L(u(t),u'(t),t)dt$ tends to be minimized. From that it's a mathematical theorem that if $L(u,\dot{u},t)=L(u,\dot{u},t+t_0)$ for all $t_0$, then energy is conserved. However then your question becomes, "why does nature tend to minimize the action" or equivalently, "why must we use a function like $L$?" To which one must appeal to experiment! There are no proofs of energy conservation just as there are no proofs of Newton's laws!

Answer 2

Derivation that applied work as defined above results, for a particle moving along a straight line, in a change in its kinetic energy (I hope it is not too complex to understand):

In the case the resultant force $F$ is constant in both magnitude and direction, and parallel to the velocity of the particle, the particle is moving with constant acceleration a along a straight line. The relation between the net force and the acceleration is given by the equation $F = ma$ (Newton's second law), and the particle displacement $s$ can be expressed by the equation $s = \frac{v_2^2 - v_1^2}{2a}$ which follows from $v_2^2 = v_1^2 + 2as$.

The work of the net force is calculated as the product of its magnitude and the particle displacement. Substituting the above equations, one obtains: $W = Fs = mas = ma \left(\frac{v_2^2 - v_1^2}{2a}\right) = \frac{mv_2^2}{2} - \frac{mv_1^2}{2} = \Delta {E_k}$

In the general case of rectilinear motion, when the net force $F$ is not constant in magnitude, but is constant in direction, and parallel to the velocity of the particle, the work must be integrated along the path of the particle:

$W = \int_{t_1}^{t_2} \mathbf{F}\cdot \mathbf{v}dt = \int_{t_1}^{t_2} F \,v dt = \int_{t_1}^{t_2} ma \,v dt = m \int_{t_1}^{t_2} v \,{dv \over dt}\,dt = m \int_{v_1}^{v_2} v\,dv = \tfrac12 m (v_2^2 - v_1^2)$ .