6.8 The conservation of mechanical energy

THE CONSERVATION OF MECHANICAL ENERGY

For simplicity, we demonstrate this important principle for one-dimensional motion. Assume that a body undergoes displacement \(\Delta x\) under the action of a conservative force \(F\). Then from the Work-Energy theorem, we have,
\(\Delta K=F(x) \Delta x\)
If the force is conservative, the potential energy function \(V(x)\) can be defined such that
\(-\Delta V=F(x) \Delta x\)
The above equations imply that
\(
\begin{array}{l}
\Delta K+\Delta V=0 \\
\Delta(K+V)=0 \dots(6.10)
\end{array}
\)
which means that \(K+V\), the sum of the kinetic and potential energies of the body is a constant. Over the whole path, \(x_{i}\) to \(x_{f}\), this means that
\(
K_{i}+V\left(x_{i}\right)=K_{f}+V\left(x_{f}\right) \dots (6.11)
\)
The quantity \(K+V(x)\), is called the total mechanical energy of the system. Individually the kinetic energy \(K\) and the potential energy \(V(x)\) may vary from point to point, but the sum is a constant. 
Let us consider some of the definitions of a conservative force.

• A force \(F(x)\) is conservative if it can be derived from a scalar quantity \(V(x)\) by the relation given by \(\Delta V=-F(x) \Delta x\).
• The work done by the conservative force depends only on the endpoints. This can be seen from the relation,
  \(W=K_{f}-K_{i}=V\left(x_{i}\right)-V\left(x_{f}\right)\), which depends on the end points.
• A third definition states that the work done by this force in a closed path is zero. This is once again apparent from Equation (6.8) since \(x_{i}=x_{f}\).

Thus, the principle of conservation of total mechanical energy can be stated as, the total mechanical energy of a system is conserved if the forces, doing work on it, are conservative.

Important Points:

• The sum of the kinetic energy and the potential energy is called the total mechanical energy. The total mechanical energy of a system remains constant if the internal forces are conservative and the external forces do no work. This is called the principle of conservation of energy.
• If nonconservative internal forces operate within the system, or external forces do work on the system, the mechanical energy changes as the configuration changes. According to the work-energy theorem, the work done by all the forces equals the change in the kinetic energy. 

The above discussion can be made more concrete by considering the example of the gravitational force once again and that of the spring force in the next section. Fig. 6.5 depicts a ball of mass \(m\) being dropped from a cliff of height \(H\).

The total mechanical energies \(E_0, E_h\), and \(E_H\) of the ball at the indicated heights zero (ground level), \(h\) and \(H\), are
\(
\begin{aligned}
E_H & =m g H \dots(6.11a) \\
E_h & =m g h+\frac{1}{2} m v_h^2 \dots(6.11b) \\
E_0 & =(1 / 2) m v_f^2 \dots(6.11c)
\end{aligned}
\)
The constant force is a special case of a spatially dependent force \(F(x)\). Hence, the mechanical energy is conserved. Thus
\(
E_H=E_0
\)
or, \(\quad m g H=\frac{1}{2} m v_f^2\)
\(
v_f=\sqrt{2 g H}
\)
a result that was obtained in section 3.7 for a freely falling body.
Further,
\(
E_H=E_h
\)
which implies,
\(
v_{\mathrm{h}}^2=2 g(H-h) \dots(6.11d)
\)
and is a familiar result from kinematics.
At the height \(H\), the energy is purely potential. It is partially converted to kinetic at height \(h\) and is fully kinetic at ground level. This illustrates the conservation of mechanical energy.

Example 6.10: A bob of mass \(m\) is suspended by a light string of length \(L\). It is imparted a horizontal velocity \(v_0\) at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point, C. This is shown in Fig. 6.6. Obtain an expression for (i) \(v_o\); (ii) the speeds at points \(\mathrm{B}\) and \(\mathrm{C}\); (iii) the ratio of the kinetic energies \(\left(K_B / K_C\right)\) at \(\mathrm{B}\) and \(\mathrm{C}\). Comment on the nature of the trajectory of the bob after it reaches the point \(C\).

Answer: (i) There are two external forces on the bob: gravity and the tension \((T)\) in the string. The latter does no work since the displacement of the bob is always normal to the string. The potential energy of the bob is thus associated with the gravitational force only. The total mechanical energy \(E\) of the system is conserved. We take the potential energy of the system to be zero at the lowest point \(A\). Thus, at \(A\):
\(
E=\frac{1}{2} m v_0^2 \dots(6.12)
\)
\(
T_A-m g=\frac{m v_0^2}{L} \text { [Newton’s Second Law] }
\)
where \(T_A\) is the tension in the string at A. At the highest point \(C\), the string slackens, as the tension in the string \(\left(T_C\right)\) becomes zero.
Thus, at \(\mathrm{C}\)
\(
\begin{aligned}
& E=\frac{1}{2} m v_c^2+2 m g L \dots(6.13)\\
& m g=\frac{m v_c^2}{L} \quad \text { [Newton’s Second Law] } \dots(6.14)
\end{aligned}
\)
where \(v_C\) is the speed at C. From Eqs. (6.13) and (6.14)
\(
E=\frac{5}{2} m g L
\)
Equating this to the energy at A
\(
\begin{array}{ll}
& \frac{5}{2} m g L=\frac{m}{2} v_o^2 \\
\text { or, } & v_0=\sqrt{5 g L}
\end{array}
\)
(ii) It is clear from Eq. (6.14)
\(
v_C=\sqrt{g L}
\)
At \(B\), the energy is
\(
E=\frac{1}{2} m v_B^2+m g L
\)
Equating this to the energy at A and employing the result from (i), namely \(v_0^2=5 g L\),
\(
\begin{aligned}
& \frac{1}{2} m v_B^2+m g L=\frac{1}{2} m v_o^2 \\
& \quad=\frac{5}{2} m g L
\end{aligned}
\)
\(
\therefore v_B=\sqrt{3 g L}
\)
(iii) The ratio of the kinetic energies at B and C is :
\(
\frac{K_B}{K_C}=\frac{\frac{1}{2} m v_B^2}{\frac{1}{2} m v_C^2}=\frac{3}{1}
\)
At the point \(\mathrm{C}\), the string becomes slack and the velocity of the bob is horizontal and to the left. If the connecting string is cut at this instant, the bob will execute a projectile motion with horizontal projection akin to a rock kicked horizontally from the edge of a cliff. Otherwise, the bob will continue on its circular path and complete the revolution.

Example 6.11: Prove that the work done by the external forces equals the change in the mechanical energy of the system. \(W_{e x t}=E_f-E_i\)

Answer: According to the work-energy theorem, the work done by all the forces equals the change in the kinetic energy. Thus,
\(
W_c+W_{n c}+W_{e x t}=K_f-K_i
\)
where the three terms on the left denote the work done by the conservative internal forces, nonconservative internal forces and external forces.
As
\(
W_c=-\left(U_f-U_i\right)
\)
we get
\(
\begin{aligned}
W_{n c}+W_{e x t} & =\left(K_f+U_f\right)-\left(K_i+U_i\right) \\
& =E_f-E_i \dots(i)
\end{aligned}
\)
where \(E=K+U\) is the total mechanical energy.
If the internal forces are conservative but external forces also act on the system and they do work, \(W_{n c}=0\) and from (i),
\(
W_{e x t}=E_f-E_i \dots(ii)
\)

Example 6.12: Two charged particles \(A\) and \(B\) repel each other by a force \(k / r^2\), where \(k\) is a constant and \(r\) is the separation between them. The particle \(A\) is clamped to a fixed point in the lab and the particle B which has a mass \(m\), is released from rest with an initial separation \(r_0\) from \(\mathrm{A}\). Find the change in the potential energy of the two-particle system as the separation increases to a large value. What will be the speed of particle B in this situation?

Answer: The situation is shown in the figure below. Take \(A+B\) as the system. The only external force acting on the system is that needed to hold \(A\) fixed. (You can imagine the experiment being conducted in a gravity free region or the particles may be kept and allowed to move on a smooth horizontal surface, so that the normal force balances the force of gravity). This force does no work on the system because it acts on the charge \(A\) which does not move. Thus, the external forces do no work and internal forces are conservative. The total mechanical energy must, therefore, remain constant. There are two internal forces; \(F_{A B}\) acting on \(A\) and \(F_{B A}\) acting on \(B\). The force \(F_{A B}\) does no work because it acts on \(A\) which does not move. The work done by \(F_{A A}\) as the particle \(B\) is taken away is,
\(
W=\int \vec{F} \cdot d \vec{r}=\int_{r_0}^{\infty} \frac{k}{r^2} d r=\frac{k}{r_0} \dots(i)
\)

The change in the potential energy of the system is
\(
U_f-U_i=-W=-\frac{k}{r_0}
\)
As the total mechanical energy is conserved,
\(
\begin{aligned}
& K_f+U_f=K_i+U_i \\
& \text { or, } \\
& K_f=K_i-\left(U_f-U_i\right) \\
& \text { or, } \quad \frac{1}{2} m v^2=\frac{k}{r_0} \\
& \text { or, } \\
& v=\sqrt{\frac{2 k}{m r_0}} . \\
&
\end{aligned}
\)

Let us summarise the concepts developed so far in this section.

• Work done on a particle is equal to the change in its kinetic energy.
• Work done on a system by all the (external and internal) forces is equal to the change in its kinetic energy.
• A force is called conservative if the work done by it during a round trip of a system is always zero. The force of gravitation, Coulomb force, force by a spring, etc. are conservative. If the work done by it during a round trip is not zero, the force is nonconservative. Friction is an example of a nonconservative force.
• The change in the potential energy of a system corresponding to conservative internal forces is equal to the negative of the work done by these forces.
• If no external forces act (or the work done by them is zero) and the internal forces are conservative, the mechanical energy of the system remains constant. This is known as the principle of conservation of mechanical energy.
• If some of the internal forces are nonconservative, the mechanical energy of the system is not constant.
• If the internal forces are conservative, the work done by the external forces is equal to the change in mechanical energy.