We have convinced ourselves in the examples just considered how helpful it is to know a quantity not changing its numerical value (conserving it) throughout a motion.

So far we know such a quantity for one body only. But if several associated bodies are moving within a gravitational field? It is evident that we may not assume that the expression \((v^{2}/2) + gh\) remains constant for each of them, since each of the bodies is subject to the action of not only the force of gravity but also of the neighbouring bodies. Perhaps the sum of such expressions taken over the group of bodies under consideration is conserved?

We shall now show that this assumption is false. There exists a quantity conserved throughout the motion of many bodies; however, it is not equal to the sum \[\left( v^{2} + gh \right)_{\textrm{body\,1}}+ \left( v^{2} + gh \right)_{\textrm{body\,2}} + \ldots\] but rather to the sum of such expressions multiplied by the masses of the corresponding bodies; in other words, the sum \[m_{1}\left( v^{2} + gh \right)_{1}+ m_{2}\left( v^{2} + gh \right)_{2}+ \ldots\] is conserved.

For the proof of this important law of mechanics, we turn to the following example.

Two loads are connected by a cord passing over a pulley, the large one of mass \(M\), and the small one of mass \(m\). The large load pulls the small one, and this group of two bodies will move with increasing speed.

The driving force is the difference in weight of these bodies, \(Mg - mg\). Since the masses of both bodies participate in the accelerated motion, Newton’s law for this case will be written out as follows: \[(M-m)g = (M+m)g\] Let us consider two instants during the motion and show that the sum of the expressions \((v^{2}/2) + gh\) multiplied by the corresponding masses really remains unchanged. Thus, it is required to prove the equality \[m\left( v_{2}^{2} + gh_{2} \right)+ M\left( V_{2}^{2} + gH_{2} \right) = m\left( v_{1}^{2} + gh_{1} \right)+ M\left( V_{1}^{2} + gH_{1} \right)\] Capital letters denote physical quantities characterizing the large load. The subscripts 1 and 2 refer here to the two instants which we are considering.

Since the loads are connected by a cord, \(v_{1} = V_{1}\) and \(v_{2} = V_{2}\). Using these simplifications and transferring all summands containing heights to the right-hand side, and summands with speeds to the left-hand side, we obtain: \begin{align} % \frac{m+M}{2} (v_{2}^{2}-v_{1}^{2}) & = mgh_{1}+MgH_{1}-mgh_{2}-MgH_{2}\\ & = mg(h_{1}-h_{2}) + Mg(H_{1}-H_{2}) \end{align} The differences in height of the loads are, of course, equal (but opposite in sign since one load rises and the other falls). Therefore, \[\frac{m+M}{2} (v_{2}^{2}-v_{1}^{2}) = g(M - m)s\] where \(s\) is the distance covered.

We learned in Section: Rectilinear Motion with Constant Acceleration that the difference between the squares of the speeds at the initial and end points of a segment of length \(s\) of a path traversed with acceleration \(a\) is equal to: \[v_{1}^{2}-v_{2}^{2} = 2as\] Substituting this expression in the preceding formula, we find: \[(m + M) a = (M - m) g\] But this is Newton’s law, which we have written out above for our example. With this we have proved what was required: for two bodies the sum of the expressions \((v^{2} /2) +gh\) multiplied by the corresponding masses¹ remains constant during the motion, or, as one says, is conserved, i.e. \[\left(\frac{mv^{2}}{2} + mgh\right) + \left(\frac{MV^{2}}{2} + MgH\right) = \textrm{const.}\]

For the case of a single body, this formula reduces to the one proved earlier: \[\frac{v^{2}}{2} + gh = \textrm{const.}\] Half the product of the mass by the square of the speed is called the kinetic energy \(K\): \[K = \frac{mv^{2}}{2}\] The product of the weight of a body by its height is called the potential energy \(U\) of the gravitational attraction of the body to the Earth: \[U = mgh\] We have proved that during the motion of a two-body system (and it is possible to prove the same thing for a system consisting of many bodies) the sum of the kinetic and potential energies of the bodies remains constant. In other words, an increase in the kinetic energy of a group of bodies can only occur at the expense of a decrease in the potential energy of this system (and, of course, conversely).

The law just proved is called the law of conservation of mechanical energy.

The law of conservation of mechanical energy is a very important law of nature. We have not yet shown its significance in full measure. Later, when we have become acquainted with the motion of molecules, its universality and its applicability to all natural phenomena will be evident.