6.2 Types of potential energy

1.1 Work done by area under force displacement curve

2.1 Properties of work done
2.2 Work done by a variable force

3.1 Work done by a spring force
3.2 Work done on a spring by an external force

6.1 Potential energy $\left( {\Delta U} \right)$ is negative of the work done by conservative forces.
6.2 Types of potential energy
6.3 Law of conservation of mechanical energy

The work done by a spring force $(W_S)$ when it is elongated or compressed by distance $x$, $${W_S} = - \frac{1}{2}k{x^2}$$

As we know, that the change in potential energy is negative of the work done by the conservative forces i.e. work done by spring force, in this case, we get, $$\begin{equation} \begin{aligned} \Delta U = - {W_S} \\ \Delta U = \frac{1}{2}k{x^2} \\\end{aligned} \end{equation} $$

Note: The elastic potential energy is always positive whether the spring is elongated or compressed.

Gravitational potential energy has a wide application in different physics problems.

Two masses $m_1$ and $m_2$, separated by a distance $r$. The force $(F)$ between the masses is, $$F = - G\frac{{{m_1}{m_2}}}{{{r^2}}}...(i)$$

In the above equation, negative (-ve) sign tells that the force is attractive in nature.

Now let us assume mass $m_1$ to be fixed.

So, the work done by gravitational force in displacing mass $m_2$ by a small displacement $dr$ we get, $$dW = - G\frac{{{m_1}{m_2}}}{{{r^2}}}dr...(ii)$$

As we know, that the gravitational force is conservative in nature.

Also, the change in potential energy is negative of the work done by the conservative forces.

Therefore, $$dU=-W ...(iii)$$

From equation $(ii)$ and $(iii)$ we get, $$dU = G\frac{{{m_1}{m_2}}}{{{r^2}}}dr...(ii)$$

For calculating absolute gravitational potential energy at a separation $r$ we will integrate the above equation from $\infty $ to separation $r$, therefore, $$\begin{equation} \begin{aligned} \int\limits_{{U_\infty }}^{{U_r}} {dU = \int\limits_\infty ^r {G\frac{{{m_1}{m_2}}}{{{r^2}}}} } dr \\ {U_r} - {U_\infty } = - G{m_1}{m_2}\left[ {\frac{1}{r} - \frac{1}{\infty }} \right] \\\end{aligned} \end{equation} $$

Since we took reference at infinity, so the potential energy at infinity is zero $\left( {{U_\infty } = 0} \right)$, and ${U_r} = U$. $$U = - G\frac{{{m_1}{m_2}}}{r}$$

$U = - G\frac{{{m_1}{m_2}}}{r}$, is the gravitational potential energy between two masses $m_1$ and $m_2$ which is separated by a distance $r$.

Gravitational potential energy between earth and a particle of mass $m$ at the surface of earth is, $${U_i} = - G\frac{{Mm}}{R}...(i)$$

Gravitational potential energy between the earth and a particle of mass $m$ at a height $H$ from the surface of earth is, $${U_f} = - G\frac{{Mm}}{{R + H}}...(ii)$$

Change in gravitational potential energy for the above two equations we get, $$\begin{equation} \begin{aligned} {U_f} - {U_i} = - GMm\left( {\frac{1}{{R + H}} - \frac{1}{R}} \right) \\ \Delta U = - GMm\frac{{\left( { - H} \right)}}{{R\left( {R + H} \right)}} \\ \Delta U = G\frac{{MmH}}{{{R^2}\left( {1 + \frac{H}{R}} \right)}} \\ As,\left( {g = \frac{{GM}}{{{r^2}}}} \right) \\ \Delta U = G\frac{{mgH}}{{\left( {1 + \frac{H}{R}} \right)}} \\ For,h < < R \\ \Delta U = mgH \\\end{aligned} \end{equation} $$

$\Delta U = mgH$, is the gravitational potential energy possessed by a particle of mass $m$ at height $H$ above the surface of the earth.

When two charges of magnitude $q_1$ and $q_2$ are separated by a distance $r$. So, according to Coulomb's law we get, $$F = k\frac{{{q_1}{q_2}}}{{{r^2}}}...(i)$$

Let us assume charge $q_1$ to be fixed.

So, the work done by electric force in displacing the charge $q_2$ by a small displacement $dr$ we get, $$dW = k\frac{{{q_1}{q_2}}}{{{r^2}}}dr...(ii)$$

As we know that the electric force is conservative in nature.

Also, the change in potential energy is negative of the work done by conservative forces.

Therefore, $$dU=-W...(iii)$$

From equations $(ii)$ and $(iii)$ we get, $$dU = - k\frac{{{q_1}{q_2}}}{{{r^2}}}dr$$

For calculating the absolute potential energy at a separation $r$, we will integrate the above equation from $\infty $ to separation $r$, therefore, $$\begin{equation} \begin{aligned} \int\limits_{{U_\infty }}^{{U_r}} {dU} = - \int\limits_\infty ^r {k\frac{{{q_1}{q_2}}}{{{r^2}}}dr} \\ {U_r} - {U_\infty } = k{q_1}{q_2}\left[ {\frac{1}{r} - \frac{1}{\infty }} \right] \\\end{aligned} \end{equation} $$

Since we took reference at infinity, so the potential energy at infinity is zero $\left( {{U_\infty } = 0} \right)$, and ${U_r} = U$. $$U = k\frac{{{q_1}{q_2}}}{r}$$

$U = k\frac{{{q_1}{q_2}}}{r}$, is the electric potential energy between two charges $q_1$ and $q_2$ which is separated by a distance $r$.

Question 12. A massless spring of spring constant $k$ is in its natural length $l$ when it is in the vertical position. The spring is connected to a bead, which can move freely along the horizontal rod. Find the elastic potential energy stored in the spring when the bead is displaced by a distance $x$ as shown in the figure.

Spring constant: $k$

Natural length of spring: $l$

Length of the spring when the bead is displaced by a distance $x$, $${l'} = \sqrt {l^2 + x^2}$$

Elongation in the spring $\left( {\Delta l} \right)$, $$\begin{equation} \begin{aligned} \Delta l = {l'} - l \\ \Delta l = \sqrt {{l^2} + {x^2}} - l \\\end{aligned} \end{equation} $$

Potential energy stored in the spring $\left( {\Delta U} \right)$, $$\begin{equation} \begin{aligned} \Delta U = \frac{1}{2}k{\left( {\Delta l} \right)^2} \\ \Delta U = \frac{1}{2}k{\left( {\sqrt {{l^2} + {x^2}} - l} \right)^2} \\\end{aligned} \end{equation} $$

Question 13. Find the change in potential energy when a particle of mass $m_2$ moves from initial to the final position as shown in the figure.

Solution: Initial gravitational potential energy between the masses $m_1$ and $m_2$. $${U_i} = - G\frac{{{m_1}{m_2}}}{{(R + H)}}$$

Final gravitational potential energy, $${U_f} = - G\frac{{{m_1}{m_2}}}{{(R + H)}}$$

Change in gravitational potential energy is given by, $$\begin{equation} \begin{aligned} \Delta U = {U_f} - {U_i} \\ \Delta U = - G\frac{{{m_1}{m_2}}}{{(R + H)}} - \left( { - G\frac{{{m_1}{m_2}}}{{(R + H)}}} \right) \\ \Delta U = 0 \\\end{aligned} \end{equation} $$

Question 14. Find the change in potential energy when charge $q_3$ is shifted from vertices $C$ to $D$, as shown in the figure. (Given ${q_1} = 1\mu C$, ${q_2} = 2\mu C$ and ${q_3} = 10\mu C$, $a=3m$ and $b=4m$).

Solution: For three charges the number of possible combination pairs for potential energy is ${}^3{C_2} = 3$.

Initial potential energy $\left( {{U_i}} \right)$, $${U_i} = k\frac{{{q_1}{q_2}}}{a} + k\frac{{{q_2}{q_3}}}{b} + k\frac{{{q_3}{q_1}}}{{\sqrt {{a^2} + {b^2}} }}$$

Final potential energy $\left( {{U_f}} \right)$, $${U_f} = k\frac{{{q_1}{q_2}}}{a} + k\frac{{{q_2}{q_3}}}{{\sqrt {{a^2} + {b^2}} }} + k\frac{{{q_3}{q_1}}}{b}$$

Change in potential energy $\left( {\Delta U} \right)$, $$\begin{equation} \begin{aligned} \Delta U = {U_f} - {U_i} \\ \Delta U = \left( {k\frac{{{q_1}{q_2}}}{a} + k\frac{{{q_2}{q_3}}}{b} + k\frac{{{q_3}{q_1}}}{{\sqrt {{a^2} + {b^2}} }}} \right) - \left( {k\frac{{{q_1}{q_2}}}{a} + k\frac{{{q_2}{q_3}}}{{\sqrt {{a^2} + {b^2}} }} + k\frac{{{q_3}{q_1}}}{a}} \right) \\ \Delta U = k\left( {\frac{{{q_3}{q_1} - {q_2}{q_3}}}{{\sqrt {{a^2} + {b^2}} }}} \right) + k\left( {\frac{{{q_2}{q_3} - {q_3}{q_1}}}{b}} \right) \\ \Delta U = k\left( {\frac{{10 - 20}}{5} - \frac{{20 - 10}}{4}} \right) \\ \Delta U = k\left( { - 4.5} \right) \\ \Delta U = -4.05 \times {10^{10}}J\quad \quad (As,k = 9 \times {10^9}N{m^2}/{C^2}) \\\end{aligned} \end{equation} $$