7.2: Conservation of energy
For certain forces, so called conservative forces , conservation of energy holds. To understand what is meant by conservation of energy, consider a ball \(i\) rolling back and forth in a half pipe under the influence of gravity like in Figure 7.3. The ball rolls up one of the ramps, stops at a maximum height, rolls downward and gains its maximum speed at the lowest point after which this sequence repeats itself. If there is no friction, the ball will continue rolling back and forth indefinitely. Apparently the system has a kind of ’memory’, represented by a quantity that is always constant.
Let us assume that there is a quantity called the internal energy \(U_{i, \text { tot }}\) which is constant (conserved) and can be defined as the sum of the kinetic energy and a potential energy function \(V_{i}(\overrightarrow{\boldsymbol{r}})\).
\[T_{i}+V_{i}(\overrightarrow{\boldsymbol{r}}) \equiv U_{i, \text { tot }}=\text { constant } \tag{7.12} \label{7.12}\]
This equation is called the law of conservation of energy. We now need to derive when it holds and what are the properties of the potential energy.
If the total internal energy \(U_{i, \text { tot }}\) is constant, then always \(\Delta U_{i, \text { tot }}=0\). When the point mass \(i\) moves from position \(\overrightarrow{\boldsymbol{r}}_{1}\) to position \(\overrightarrow{\boldsymbol{r}}_{2}\) under the influence of a force \(\overrightarrow{\boldsymbol{F}}_{i j}\), this implies that the change in its kinetic and potential energy are:
\[\Delta T_{i, \overrightarrow{\boldsymbol{r}}_{1} \rightarrow \overrightarrow{\boldsymbol{r}}_{2}}+\Delta V_{i j, \overrightarrow{\boldsymbol{r}}_{1} \rightarrow \overrightarrow{\boldsymbol{r}}_{2}}=\Delta U_{i, \text { tot }} \equiv 0 \tag{7.13} \label{7.13}\]
From the principle of work and energy, Equation 7.1, we have \(\Delta T_{i, \vec{r}_{1} \rightarrow \vec{r}_{2}}=\) \(W_{i j, \overrightarrow{\boldsymbol{r}}_{1} \rightarrow \overrightarrow{\boldsymbol{r}}_{2}}\). By combining this with Equation 7.13 we find the equation for potential energy.
The change in the potential energy of a point mass \(i\) is the negative of the work done by a conservative force \(\overrightarrow{\boldsymbol{F}}_{i j, c}\) on the point mass.
\[\Delta V_{i j, \overrightarrow{\boldsymbol{r}}_{1} \rightarrow \overrightarrow{\boldsymbol{r}}_{2}}=-W_{i j, \overrightarrow{\boldsymbol{r}}_{1} \rightarrow \overrightarrow{\boldsymbol{r}}_{2}} \tag{7.14} \label{7.14}\]
By combining Equation 7.14 with the definition of work Equation 7.2 we find that the potential energy obeys:
\[V_{i j}\left(\overrightarrow{\boldsymbol{r}}_{2}\right)-V_{i j}\left(\overrightarrow{\boldsymbol{r}}_{1}\right)=-\int_{\overrightarrow{\boldsymbol{r}_{\mathbf{1}}}}^{\overrightarrow{\boldsymbol{r}_{\mathbf{2}}}} \overrightarrow{\boldsymbol{F}}_{i j, c}(\overrightarrow{\boldsymbol{r}}) \cdot \mathrm{d} \overrightarrow{\boldsymbol{r}} \tag{7.15} \label{7.15}\]
However, not all forces obey Equation 7.15. If the point mass moves along a closed path, we have \(\overrightarrow{\boldsymbol{r}}_{2}=\overrightarrow{\boldsymbol{r}}_{1}\). If we substitute this in Equation 7.15 we find that only forces that satisfy the following equation are conservative forces:
The work done by a conservative force \(\overrightarrow{\boldsymbol{F}}_{i j, c}\) on a point mass along any closed path is zero.
\[W_{i j, s_{1} \rightarrow s_{1}}=\oint_{\overrightarrow{\boldsymbol{r}_{\mathbf{1}}}}^{\overrightarrow{\boldsymbol{r}_{\mathbf{1}}}} \overrightarrow{\boldsymbol{F}}_{i j, c}(\overrightarrow{\boldsymbol{r}}) \cdot \mathrm{d} \overrightarrow{\boldsymbol{r}}=0 \tag{7.16} \label{7.16}\]
According to this definition, the gravitational force is conservative, since it follows from Equation 7.6 that if \(\overrightarrow{\boldsymbol{r}}_{2}=\overrightarrow{\boldsymbol{r}}_{1}\), we have \(y_{2}=y_{1}\) and \(W_{i j, s_{1} \rightarrow s_{1}}=0\), irrespective of the path.
7.2.1 Force from potential energy
Interestingly, once the potential energy function \(V_{i j}\left(\overrightarrow{\boldsymbol{r}}_{i}\right)\) of a conservative force is known, it can be used to determine the force field vector by taking its gradient as follows:
\[\overrightarrow{\boldsymbol{F}}_{i j, c}(\overrightarrow{\boldsymbol{r}})=-\overrightarrow{\boldsymbol{\nabla}} V_{i j}(\overrightarrow{\boldsymbol{r}}) \tag{7.17} \label{7.17}\]
By combining Equation 7.4 and (7.14), we obtain for the change in potential energy when the point mass moves from \(\overrightarrow{\boldsymbol{r}}_{1}\) to \(\overrightarrow{\boldsymbol{r}}_{2}\) :
\[\Delta V_{i j, \overrightarrow{\boldsymbol{r}}_{1} \rightarrow \overrightarrow{\boldsymbol{r}}_{2}}=-\int_{s_{1}}^{s_{2}} F_{i j, t} \mathrm{~d} s \tag{7.18} \label{7.18}\]
If one chooses the path curve to be parallel to the \(x\) axis, \(\mathrm{d} s\) becomes \(\mathrm{d} x\) and by substituting \(s_{1}=x\) and \(s_{2}=x+\Delta x\) in Equation 7.18, in the limit \(\Delta x \rightarrow 0\) we obtain:
\[\begin{align} V_{i j}(x+\Delta x)-V_{i j}(x) & =-F_{i j, x} \Delta x \tag{7.19} \label{7.19}\\[4pt] \lim _{\Delta x \rightarrow 0} \frac{V_{i j}(x+\Delta x)-V_{i j}(x)}{\Delta x} & =-F_{i j, x} \tag{7.20} \label{7.20}\\[4pt] F_{i j, x} & =-\frac{\partial V_{i j}(\overrightarrow{\boldsymbol{r}})}{\partial x} \tag{7.21} \label{7.21}\end{align}\]
A similar procedure as in Equation 7.21 can be used to obtain the other components of the force vector, by taking the path curve parallel to \(y\) and \(z\) axes, such that the total force vector at position \(\overrightarrow{\boldsymbol{r}}\) can be constructed:
\[\begin{align} \overrightarrow{\boldsymbol{F}}_{i j, c}(\overrightarrow{\boldsymbol{r}}) & =F_{i j, x} \hat{\boldsymbol{\imath}}+F_{i j, y} \hat{\boldsymbol{\jmath}}+F_{i j, z} \hat{\boldsymbol{k}} \tag{7.22} \label{7.22}\\[4pt] & =-\frac{\partial V_{i j}}{\partial x} \hat{\boldsymbol{\imath}}-\frac{\partial V_{i j}}{\partial y} \hat{\boldsymbol{\jmath}}-\frac{\partial V_{i j}}{\partial z} \hat{\boldsymbol{k}} \tag{7.23} \label{7.23}\\[4pt] & =-\overrightarrow{\boldsymbol{\nabla}} V_{i j}(\overrightarrow{\boldsymbol{r}}) \tag{7.24} \label{7.24}\end{align}\]
Here we used that the del or nabla operator \(\overrightarrow{\boldsymbol{\nabla}}\) in Cartesian coordinates is \(\overrightarrow{\boldsymbol{\nabla}}=\frac{\partial}{\partial x} \hat{\boldsymbol{\imath}}+\frac{\partial}{\partial y} \hat{\boldsymbol{\jmath}}+\frac{\partial}{\partial z} \hat{\boldsymbol{k}} . \overrightarrow{\boldsymbol{\nabla}} V\) is the gradient of the scalar potential energy function. Essentially, this gradient shows that the force on a point mass always points downward along the steepest ’downhill’ potential direction, with a magnitude equal to the slope of the potential energy function along that direction.
7.2.2 Heat and energy conservation*
Interestingly, all fundamental forces of nature are conservative forces, which means that energy can never be lost and remains constant. Nevertheless, besides increasing the kinetic energy of large point-masses, forces like friction can also increase the random velocity of the atoms in a material. The energy associated with this random motion of the atoms is called heat \(Q\), and leads to temperature increases. Unfortunately, since the motion is random, it is not possible to fully reuse heat to perform work, and can thus be considered
as (partly) lost energy. Forces that generate heat are therefore also called non-conservative forces or dissipative forces.
For example when a block slides off a ramp while friction forces act on it, the temperature of the block and the ramp increase due to the higher average velocity of the atoms in the materials. Non-conservative forces like friction usually generate heat \((\Delta Q>0)\). To distinguish the potential and kinetic energy stored in a closed system from heat energy and external sources or sinks of energy, the sum of kinetic and potential energy \(T+V=U_{i}\) of a point mass or system of point masses is called its internal energy. Internal energy can reduce via generation of heat \(Q\), and can also reduce by net work \(W_{\text {ext }}\) performed by the internal system on an external system. If the effect of heat and work is included, the energy conservation equation becomes:
\[T+V+Q+W_{\text {ext }}=E_{\text {tot }}=\text { constant } \tag{7.25} \label{7.25}\]
This equation, that is depicted schematically in Figure 7.4 is also called the first law of thermodynamics, and is discussed in more detail in thermodynamics textbooks. If we only look at changes in energy, the law can be reformulated into:
\[\Delta T+\Delta V+\Delta Q+\Delta W_{\text {ext }}=0 \tag{7.26} \label{7.26}\]
The equation states that total energy \(E_{\text {tot }}\) is conserved and can be interchanged between kinetic and potential energy, or heat energy \(Q\) or work \(W_{\text {ext }}\) on external systems. We note that there are always dissipative, frictional forces, which will continue to convert internal energy into heat as long as the system keeps moving. Therefore, unless there are external sources of work or heat \(\left(\Delta Q+\Delta W_{\text {ext }}<0\right)\), the system will eventually stop moving once the internal energy is all converted into heat. As a consequence it is impossible to create so-called perpetuum mobiles: mechanisms that continue moving forever without energy supply.
7.2.3 Power
The work \(W\) of a force on a point mass can be divided by the time interval over which this work is done to obtain the average power generated by the force. When the time interval tends to zero we obtain that the (instantaneous) power generated by a force is:
\[P=\frac{\mathrm{d} W}{\mathrm{~d} t} \tag{7.27} \label{7.27}\]
From Equation 7.2 it can be seen that \(\mathrm{d} W=\overrightarrow{\boldsymbol{F}} \cdot \mathrm{d} \overrightarrow{\boldsymbol{r}}\), from which we find that the rate at which power is generated equals the inner product of the force and velocity vectors:
\[P=\frac{\mathrm{d} W}{\mathrm{~d} t}=\overrightarrow{\boldsymbol{F}} \cdot \frac{\mathrm{d} \overrightarrow{\boldsymbol{r}}}{\mathrm{d} t}=\overrightarrow{\boldsymbol{F}} \cdot \overrightarrow{\boldsymbol{v}} \tag{7.28} \label{7.28}\]
Inversely, the total work done by a force can be obtained by integrating the power over time:
\[W=\int P \mathrm{~d} t=\int \overrightarrow{\boldsymbol{F}} \cdot \overrightarrow{\boldsymbol{v}} \mathrm{d} t \tag{7.29} \label{7.29}\]
7.2.4 Efficiency
Let’s say we want to determine the efficiency of a car engine. The external input energy is the fuel consumed by the car engine which is converted into kinetic, potential and heat energy and is thus equal to \(\Delta W_{\text {in }}=\Delta T+\Delta V+\Delta Q\). The useful output energy is the kinetic energy and potential energy increase of the car \(\Delta T+\Delta V\), that it uses to accelerate or to drive uphill. The non-useful energy is the heat that is generated in the car \(\Delta Q\).
The efficiency of the car engine is then defined as:
\[\eta=\frac{\text { Useful output energy }}{\text { Total input energy }} \times 100 \%=\frac{\Delta T+\Delta V}{\Delta T+\Delta V+\Delta Q} \times 100 \% \tag{7.30} \label{7.30}\]
By taking the time derivative of the numerator and denominator of this equation one obtains the instantaneous efficiency:
\[\eta=\frac{P_{\mathrm{out}}}{P_{\mathrm{in}}} \times 100 \%=\frac{\text { Useful output power }}{\text { Total input power }} \times 100 \% \tag{7.31} \label{7.31}\]
The equation for efficiency can depend on the type of application. In a car, heat generation \(Q\) is not useful output energy, but in a heating appliance it is.
A block with mass \(m\) slides from a hill. It starts from rest at a height \(y=h\) and reaches a final velocity \(v_{2}\) at height \(y=0\). Determine the efficiency of the conversion from potential to kinetic energy of the block.
Solution
This problem is similar to the example in the previous chapter, see Figure 7.5.
In this case, the input energy is equal to the reduction in potential energy of the block \(-\Delta V=m g h\). The change in kinetic energy \(\Delta T=\frac{1}{2} m v_{2}^{2}\) of the block is the useful energy. Therefore, using Equation 7.30, we obtain:
\[\eta=\frac{\Delta T}{-\Delta V} \times 100 \%=\frac{\frac{1}{2} m v_{2}^{2}}{m g h} \times 100 \% \tag{7.32} \label{7.32}\]
As a challenge, using Eqs. (6.21)-(6.23), show that for a kinetic friction coefficient \(\mu_{k}\) and slope angle \(\alpha\) the efficiency is:
\[\eta=\left(1-\frac{\mu_{k}}{\tan \alpha}\right) \times 100 \% \tag{7.33} \label{7.33}\]
7.2.5 Conservation of energy for a system
Let us now consider the situation when there are multiple point masses \(i\) in a system, each with a potential energy function \(V_{i}\). Instead of considering them separately, it is sometimes useful to treat them together, e.g. when the forces they exert on each other are not well known. In that case we can sum all kinetic and potential energies from Equation 7.12 at times \(t=t_{1}\) and \(t=t_{2}\) to obtain:
\[\sum_{i} T_{i}\left(t_{1}\right)+\sum_{i} V_{i}\left(t_{1}\right)=\sum_{i} T_{i}\left(t_{2}\right)+\sum_{i} V_{i}\left(t_{2}\right)=\sum_{i} E_{i, \mathrm{tot}} \tag{7.34} \label{7.34}\]
The total energy of the system of point masses is conserved, and is equal to \(\sum_{i} E_{i, \text { tot }}\) at all times.