Based on a geometric demonstration by Johann Bernoulli, we show that the principles underlying the work–energy theorem are more fundamental than those discussed in the standard derivations. By introducing the notion of a mechanical Bernoulli process—a systematic process for destroying velocity and measuring work—we derive the relationship between the amount of work and the square of the velocity solely from the superposition of motion and the law of inertia.

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If we let the work sinks (sources) be infinitesimal objects, that is, infinitesimal forces and/or displacements, then the velocity of the standard mass can be destroyed (created) in a continuous fashion. In this case, the work quantum

*w*is an infinitesimal quantity and the values of the net work $(W=w,2w,3w,\u2026)$ can be approximated by a continuum. As a further generalization, we could destroy the velocity in steps of unequal work using nonuniform obstacles.14.

In the language of vector analysis, the decomposition (superposition) of motion says that the net motion $r(t)$ can be analyzed as a linear superposition of independent component motions: $r(t)=r\u22a5(t)+r\u2225(t),$ or $r(t)=x(t)i+y(t)j.$ In dynamical terms, the component motions are independent because $x(t)$ depends only on $Fx\u200a(via\u200aFx=mdvx/dt),$ and $y(t)$ depends only on $Fy\u200a(via\u200aFy=mdvy/dt).$ It follows that $\u222b\u200aFxdx$ depends only on the

*x*motion. The independence of $x(t)$ and $y(t)$ does not hold in the relativistic limit because in the relativistic equation of motion, $Fx=md(\gamma vx)/dt,$ the factor $\gamma \u2261[1\u2212(v/c)2]\u22121/2$ depends on $vx$ and $vy.$ It follows that, in a relativistic Bernoulli process, the work done by a particle in displacing an obstacle in the*x*direction depends on the motion of the particle in both the*x*and*y*directions. Indeed, work is not proportional to the square of the velocity in relativistic mechanics.15.

The mechanical Bernoulli process can be reversed. For this inverse process, each obstacle functions as a work source (mass launcher). Each work source changes the parallel component of the velocity from 0 to

*u*, thereby supplying one unit*w*of work. The increase in speed during the*n*th collision, from $vn$ to $vn+1,$ is given by the recursion relation, $vn+12=vn2+u2.$16.

H. E. Stanley,

*Introduction to Phase Transitions and Critical Phenomena*(Oxford U.P., University, New York, 1971), pp. 176–177.17.

Strictly speaking, we have derived the functional form, $W(v)\u223cv2,$ for discrete values of $v$ given by $v=\u221aNu,$ where $N=1,2,3,\u2026$ denotes the number of obstacles. If we assume that $W(v)$ is a continuous and increasing function, then we can argue that the relation, $W(v)\u223cv2,$ is valid for all real values of $v.$ Alternatively, if we use infinitesimal obstacles (infinitesimal

*u*and*w*), then the speed $v=\u221aNu$ and the work $W=Nw$ are effectively continuous variables.18.

To prove the impulse-momentum theorem, we treat each obstacle as an impulse source or impulse sink. Instead of associating one unit of scalar work

*w*with the speed*u*, we associate one unit of vector impulse**i**with the vector**u**. Each obstacle supplies the standard impulse**i**to destroy the parallel velocity**u**. Note that**u**is equal to $vn\u2212vn+1$ in a Bernoulli process. The net work*W*is determined by the number of overcome obstacles, but the net impulse**I**depends on both the number and the orientation of the overcome obstacles. Each 1 that appears in the velocity diagrams of Fig. 6 represents one unit of impulse. The velocity diagrams show that the vector sum of these unit impulse vectors is equal to the initial velocity vector. Thus, the net impulse is equal to the total change in the velocity.19.

If we introduce the standard force function $f(x)$ to characterize the resistance of the standard obstacle, then the relation between the standard parameters

*w*,*m*, and*u*is determined by the Newtonian equations, $\u222b\u200afdx=\u2212w$ and $\u222b\u200afdt=\u2212mu.$ These two equations, which for constant (or average) force reduce to $fd=\u2212w$ and $ft=\u2212mu,$ can be interpreted as the analytical meaning of the standard parameters*m*,*u*,*d*, and*w*that appear in the operational definition of work. These force equations imply the work equation $w=mu2/2.$20.

In essence, $w=mu2/2$ is the definition of the work unit, while $W(v)=mv2/2$ is the theorem on overcoming obstacles. The relation $w=mu2/2$ is valid only for the special reference (unit) values of

*m*,*u*, and*w*that parametrize the standard interaction between the standard mass and one standard obstacle. In the Bernoulli picture, we cannot assume that a similar relation holds for the general values of*m*, $v,$ and*W*that characterize the nonstandard interaction involving multiple mass-obstacle collisions. Note that the relation $w=mu2/2$ immediately converts Eqs. (6) and (11) into $1/2mvn2\u22121/2mvn+12=w$ and $W(v)=1/2mv2,$ respectively.21.

It is important to distinguish between the different dynamical underpinnings of $W\u221dmv2$ (Sec. IV) and $W=mv2/2$ (Sec. V). Whereas Bernoullian dynamics $(N=J2)$ of the whole process (overcoming

*N*obstacles) is sufficient to prove the proportionality, $W\u221dmv2,$ the proof of the equality, $W=mv2/2,$ requires Bernoullian dynamics $(N=J2)$ of the whole plus Newtonian dynamics $(w=mu2/2)$ of the parts (overcoming one obstacle). Our proof of $W\u221dmv2$ hinges on the law of inertia (Newton’s first law), while our proof of $W=mv2/2$ requires both the law of inertia and the general law of motion (Newton’s second law). In short, we get a little (factor of 1/2) from a lot (Newton’s second law).
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