Potential Energy
Potential energy is defined only for conservative forces. In the space occupied by conservative forces every point is associated with certain energy which is called the energy of position or potential energy. Potential energy generally are of three types : Elastic potential energy, Electric potential energy and Gravitational potential energy.
(1) Change in potential energy : Change in potential energy between any two points is defined in the terms of the work done by the associated conservative force in displacing the particle between these two points without any change in kinetic energy.
\({{U}_{2}}-{{U}_{1}}=-\int_{\,{{r}_{1}}}^{\,{{r}_{2}}}{\vec{F}.\,d\vec{r}=-W}\) …(i)
We can define a unique value of potential energy only by assigning some arbitrary value to a fixed point called the reference point. Whenever and wherever possible, we take the reference point at infinity and assume potential energy to be zero there, i.e. if we take \({{r}_{1}}=\infty \,\, and \,\,{{r}_{2}}=r\) then from equation (i)
\(U=-\int_{\,\infty }^{\,r}{\vec{F}.\,d\vec{r}=-W}\)
In case of conservative force (field) potential energy is equal to negative of work done by conservative force in shifting the body from reference position to given position.
This is why, in shifting a particle in a conservative field (say gravitational or electric), if the particle moves opposite to the field, work done by the field will be negative and so change in potential energy will be positive i.e. potential energy will increase. When the particle moves in the direction of field, work will be positive and change in potential energy will be negative i.e. potential energy will decrease.
(2) Three dimensional formula for potential energy: For only conservative fields \(\vec{F}\) equals the negative gradient \((-\vec{\nabla })\) of the potential energy.
So \(\vec{F}=-\vec{\nabla }U\) (\(\vec{\nabla }\)read as Del operator or Nabla operator and \(\vec{\nabla }=\frac{\partial }{\partial x}\hat{i}+\frac{\partial }{\partial y}\hat{j}+\frac{\partial }{\partial z}\hat{k}\))
\(\vec{F}=-\left[ \frac{\partial U}{\partial x}\hat{i}+\frac{\partial U}{\partial y}\hat{j}+\frac{\partial U}{\partial z}\hat{k} \right]\)
where,
\(\frac{\partial U}{\partial x}=\)Partial derivative of U w.r.t. x (keeping y and z constant)
\(\frac{\partial U}{\partial y}=\) Partial derivative of U w.r.t. y (keeping x and z constant)
\(\frac{dU}{dz}=\)Partial derivative of U w.r.t. z (keeping x and y constant)
(3) Potential energy curve : A graph plotted between the potential energy of a particle and its displacement from the centre of force is called
potential energy curve.
Figure shows a graph of potential energy function U(x) for one dimensional motion.
(4) Nature of force :
(i) Attractive force :
On increasing x, if U increases, \(\frac{dU}{dx}=\text{positive}\) , then F is in negative direction i.e. force is attractive in nature.
In graph this is represented in region BC.
(ii) Repulsive force : On increasing x, if U decreases, \(\frac{dU}{dx}=\text{negative}\), then F is in positive direction i.e. force is repulsive in nature.
In graph this is represented in region AB.
(iii) Zero force : On increasing x, if U does not change, \(\frac{dU}{dx}=0\) then F is zeroi.e. no force works on the particle.
Point B, C and D represents the point of zero force or these points can be termed as position of equilibrium.
(5) Types of equilibrium : If net force acting on a particle is zero, it is said to be in equilibrium.
For equilibrium \(\frac{dU}{dx}=0\), but the equilibrium of particle can be of three types :
Elastic Potential Energy
(1) Restoring force and spring constant : When a spring is stretched or compressed from its normal position (x = 0) by a small distance x, then a restoring force is produced in the spring to bring it to the normal position.
According to Hooke’s law this restoring force is proportional to the displacement x and its direction is always opposite to the displacement.
i.e.\(\overrightarrow{F}\propto -\overrightarrow{x\,}\)
or \(\overrightarrow{F}=-\,k\,\overrightarrow{x\,}\) …(i)
where k is called spring constant.
If x = 1, F = k (Numerically)
or k = F
Hence spring constant is numerically equal to force required to produce unit displacement (compression or extension) in the spring. If required force is more, then spring is said to be more stiff and vice-versa.
Actually k is a measure of the stiffness/softness of the spring.
Dimension : As \(k=\frac{F}{x}\)
\([k]=\frac{[F]}{[x]}=\frac{[ML{{T}^{-2}}]}{L}=[M{{T}^{-2}}]\)
Units : S.I. unit Newton/metre, C.G.S unit Dyne/cm.
Note: Dimension of force constant is similar to surface tension.
(2) Expression for elastic potential energy : When a spring is stretched or compressed from its normal position (x = 0), work has to be done by external force against restoring force.\({{\overrightarrow{F}}_{\text{ext}}}=-{{\overrightarrow{F}}_{restoring}}=k\overrightarrow{x}\)
Let the spring is further stretched through the distance dx, then work done
\(dW={{\overrightarrow{F}}_{\text{ext}}}.\,d\overrightarrow{x}={{F}_{\text{ext}}}.\,dx\cos {{0}^{o}}=kx\,dx\) [As cos 0o = 1]
Therefore total work done to stretch the spring through a distance x from its mean position is given by
\(W=\int_{0}^{x}{\,dW}=\int_{0}^{x}{\,\,kx\,dx}=k\,\left[ \frac{{{x}^{2}}}{2} \right]_{0}^{x}=\frac{1}{2}k{{x}^{2}}\)
This work done is stored as the potential energy in the stretched spring.
\(\therefore\) Elastic potential energy \(U=\frac{1}{2}k{{x}^{2}}\)
\(U=\frac{1}{2}Fx\) \(\left[ \text{As }k=\frac{F}{x} \right]\)
\(U=\frac{{{F}^{2}}}{2k}\) \(\left[ \text{As }x=\frac{F}{k} \right]\)
\(\therefore\)Elastic potential energy \(U=\frac{1}{2}k{{x}^{2}}=\frac{1}{2}Fx=\frac{{{F}^{2}}}{2k}\)
Note: If spring is stretched from initial position \({{x}_{1}}\)to final position \({{x}_{2}}\) then work done = Increment in elastic potential energy \(=\frac{1}{2}k(x_{2}^{2}-x_{1}^{2})\)
Work done by the spring-force on the block in various situation are shown in the following table:
Table : Work done for spring
(3) Energy graph for a spring : If the mass attached with spring performs simple harmonic motion about its mean position then its potential energy at any position (x) can be given by
\(U=\frac{1}{2}k{{x}^{2}}\) …(i)
So for the extreme position
\(U=\frac{1}{2}k{{a}^{2}}\) [As x = ± a for extreme]
This is maximum potential energy or the total energy of mass.
\(\therefore\) Total energy \(E=\frac{1}{2}k{{a}^{2}}\) …(ii)
[Because velocity of mass is zero at extreme position]
\(\therefore\) \(K=\frac{1}{2}m{{v}^{2}}=0\)]
Now kinetic energy at any position
\(K=E-U=\frac{1}{2}k\,{{a}^{2}}-\frac{1}{2}k\,{{x}^{2}}\)
\(K=\frac{1}{2}k({{a}^{2}}-{{x}^{2}})\) …(iii)
From the above formula we can check that
\({{U}_{\text{max}}}=\frac{1}{2}k{{a}^{2}}\) [At extreme x = ± a]
and \({{U}_{\text{min}}}=0\) [At mean x = 0]
\({{K}_{\text{max}}}=\frac{1}{2}k{{a}^{2}}\) [At mean x = 0]
and \({{K}_{\text{min}}}=0\) [At extreme x = ± a]
\(E=\frac{1}{2}k{{a}^{2}}=\) constant (at all positions)
It means kinetic energy and potential energy changes parabolically w.r.t. position but total energy remain always constant irrespective to position of the mass.
Electrical Potential Energy
It is the energy associated with state of separation between charged particles that interact via electric force. For two point charge \({{q}_{1}}\) and \({{q}_{2}}\), separated by distance r.
\(U=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}_{1}}{{q}_{2}}}{r}\)
While for a point charge q at a point in an electric field where the potential is V
U = qV
As charge can be positive or negative, electric potential energy can be positive or negative.
It is the usual form of potential energy and this is the energy associated with the state of separation between two bodies that interact via gravitational force.
For two particles of masses m1 and m2 separated by a distance r
Gravitational potential energy \(U=-\frac{G\,{{m}_{1}}{{m}_{2}}}{r}\)
(1) If a body of mass m at height h relative to surface of earth then
Gravitational potential energy \(U=\frac{mgh}{1+\frac{h}{R}}\)
Where R = radius of earth, g = acceleration due to gravity at the surface of the earth.
(2) If h << R then above formula reduces to U = mgh.
(3) If V is the gravitational potential at a point, the potential energy of a particle of mass m at that point will be
U = mV
(4) Energy height graph : When a body projected vertically upward from the ground level with some initial velocity then it possess kinetic energy but its initial potential energy is zero.
As the body moves upward its potential energy increases due to increase in height but kinetic energy decreases (due to decrease in velocity). At maximum height its kinetic energy becomes zero and potential energy maximum but through out the complete motion, total energy remains constant as shown in the figure.
A chain of length L and mass M is held on a frictionless table with (1/n)th of its length hanging over the edge.
Let \(m=\frac{M}{L}=\) mass per unit length of the chain and y is the length of the chain hanging over the edge. So the mass of the chain of length y will be ym and the force acting on it due to gravity will be mgy
The work done in pulling the dy length of the chain on the table.
dW = F(– dy) [As y is decreasing]
i.e. dW = mgy (– dy)
So the work done in pulling the hanging portion on the table.
\(W=-\int_{L/n}^{0}{mgy\,dy}=-mg\,\left[ \frac{{{y}^{2}}}{2} \right]_{L/n}^{0}=\frac{mg\,{{L}^{2}}}{2{{n}^{2}}}\)
\(\therefore\) \(W=\frac{MgL}{2{{n}^{2}}}\) [As m = M/L]
Alternative method :
If point mass m is pulled through a height h then work done W = mgh
Similarly for a chain we can consider its centre of mass at the middle point of the hanging part i.e. at a height of L/(2n) from the lower end and mass of the hanging part of chain \(=\frac{M}{n}\)
So work done to raise the centre of mass of the chain on the table is given by
\(W=\frac{M}{n}\times g\times \frac{L}{2n}\) [As W = mgh]
or \(W=\frac{MgL}{2{{n}^{2}}}\)
Taking surface of table as a reference level (zero potential energy)
Potential energy of chain when 1/nth length hanging from the edge \(=\frac{-MgL}{2{{n}^{2}}}\) Potential energy of chain when it leaves the table \(=-\frac{MgL}{2}\)Kinetic energy of chain = loss in potential energy
\(\Rightarrow\frac{1}{2}M{{v}^{2}}=\frac{MgL}{2}-\frac{MgL}{2{{n}^{2}}}\)
\(\Rightarrow\frac{1}{2}M{{v}^{2}}=\frac{MgL}{2}\left[ 1-\frac{1}{{{n}^{2}}} \right]\)
\(\therefore\) Velocity of chain \(v=\sqrt{gL\left( 1-\frac{1}{{{n}^{2}}} \right)}\)