Concept Card: Gravitational & Elastic Potential Energy & Potential Curves
1. Fundamental Concept of Potential Energy ($U$):
Potential energy is the configuration energy possessed by a system of interacting particles by virtue of their positions or internal strain relative to each other within a conservative field.
- Exclusive to Conservative Forces: Potential energy is strictly undefined for non-conservative forces (e.g., friction, viscous drag) where work depends on the path.
- Mathematical Definition:
$\Delta U = U_f – U_i = -W_{\text{conservative}} = -\int_{\vec{r}_i}^{\vec{r}_f} \vec{F}_c \cdot d\vec{r}$ - Force as Negative Gradient of Potential:
In 1D: $F_c(x) = -\frac{dU}{dx}$
In 3D: $\vec{F}_c = -\vec{\nabla}U = -\left(\frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j} + \frac{\partial U}{\partial z}\hat{k}\right)$
Physical Principle: A conservative force always points in the direction of steepest decrease of potential energy. - Arbitrary Reference Level: Only potential energy differences ($\Delta U$) are physically meaningful. Shifting the reference point by adding a constant $C$ to $U(\vec{r})$ leaves the physical force field $\vec{F} = -\vec{\nabla}U$ invariant.
2. Gravitational Potential Energy ($U_g$):
- Near Earth’s Surface (Uniform Field $\vec{g}$):
Taking ground level ($y = 0$) as reference ($U_g(0) = 0$):
$U_g(y) = mgy$
($U_g > 0$ above reference; $U_g < 0$ below reference). - Continuous / Distributed Bodies:
For an extended mass $M$ distributed over height, $U_g = M g y_{\text{cm}}$, where $y_{\text{cm}}$ is the vertical position of its center of mass.
3. Elastic Potential Energy of an Ideal Spring ($U_s$):
From Hooke’s law, $\vec{F}_s = -kx\hat{i}$. Taking the unstretched state ($x = 0$) as reference ($U_s(0) = 0$):
$$U_s(x) = -\int_{0}^{x} (-kx’)\,dx’ = \frac{1}{2}kx^2$$
- $U_s(x) \ge 0$ always: Elastic energy is strictly positive for both elongation ($x > 0$) and compression ($x < 0$).
- Spring Cutting Rule: For a uniform spring of length $L$ and constant $k$, $k \cdot L = \text{constant}$. Cutting a spring into lengths in ratio $m:n$ yields stiffer pieces: $k_1 = \left(\frac{m+n}{m}\right)k$ and $k_2 = \left(\frac{m+n}{n}\right)k$.
4. Equilibrium Classification from Potential Energy Curves ($U-x$ Graph):
At equilibrium, the net force is zero: $F = -\frac{dU}{dx} = 0$ (extrema of $U(x)$).
- Stable Equilibrium (Local Minimum): $\frac{dU}{dx} = 0 \quad \text{and} \quad \frac{d^2U}{dx^2} > 0$. Small displacement creates a restoring force. Small oscillations have angular frequency $\omega = \sqrt{\frac{1}{m}\left.\frac{d^2U}{dx^2}\right|_{\text{eq}}}$.
- Unstable Equilibrium (Local Maximum): $\frac{dU}{dx} = 0 \quad \text{and} \quad \frac{d^2U}{dx^2} < 0$. Small displacement accelerates the particle away from equilibrium.
- Neutral Equilibrium (Flat Plateau): $\frac{dU}{dx} = 0 \quad \text{and} \quad \frac{d^2U}{dx^2} = 0$.
Solved Examples
Example 1 (Chain Sliding Off a Table & Gravitational Potential Energy):
A uniform flexible chain of mass $M = 3.0\text{ kg}$ and length $L = 2.0\text{ m}$ rests on a smooth table such that a fraction $f = \frac{1}{3}$ of its length hangs over the edge. Taking $g = 10\text{ m/s}^2$:
(a) Find the initial gravitational potential energy of the chain with the tabletop as datum ($U = 0$).
(b) Calculate the work required to pull the hanging portion slowly back onto the table.
(c) If released from rest, calculate its speed when the entire chain just slips completely off the table.
Solution:
(a) The table portion is at $y = 0 \implies U = 0$.
The hanging portion has length $L_{\text{hang}} = \frac{L}{3} = \frac{2.0}{3}\text{ m}$ and mass $M_{\text{hang}} = \frac{M}{3} = 1.0\text{ kg}$.
Its center of mass lies at depth $y_{\text{cm}} = -\frac{L_{\text{hang}}}{2} = -\frac{1}{3}\text{ m}$.
$U_i = M_{\text{hang}} g y_{\text{cm}} = 1.0(10)\left(-\frac{1}{3}\right) = -\frac{10}{3}\text{ Joules} \approx -3.33\text{ J}$.
(b) $W_{\text{ext}} = U_f – U_i = 0 – \left(-\frac{10}{3}\right) = +\frac{10}{3}\text{ Joules} \approx +3.33\text{ J}$ (Formula: $W_{\text{ext}} = \frac{MgL}{2n^2}$).
(c) When the whole chain hangs vertically, its center of mass is at $y_{\text{cm}}’ = -\frac{L}{2} = -1.0\text{ m}$.
$U_f = Mg y_{\text{cm}}’ = 3.0(10)(-1.0) = -30.0\text{ Joules}$.
By conservation of energy: $0 – \frac{10}{3} = \frac{1}{2}(3.0)v^2 – 30.0 \implies 1.5 v^2 = \frac{80}{3} \implies v = \frac{4\sqrt{10}}{3} \approx 4.22\text{ m/s}$.
Example 2 (Spring Cutting & Stored Energy Comparison):
A spring of stiffness $k = 900\text{ N/m}$ and natural length $L$ is cut into two segments in length ratio $1:2$.
(a) Find the spring constants $k_1$ and $k_2$.
(b) Compare stored energies $\frac{U_1}{U_2}$ if both pieces are stretched by the same elongation $x = 2.0\text{ cm}$.
(c) Compare stored energies $\frac{U_1′}{U_2′}$ if both pieces are pulled by the same force $F = 180\text{ N}$.
Solution:
(a) $k_1 = 3k = 2700\text{ N/m}$, and $k_2 = \frac{3}{2}k = 1350\text{ N/m}$.
(b) Equal extension $x$: $U = \frac{1}{2}kx^2 \propto k \implies \frac{U_1}{U_2} = \frac{k_1}{k_2} = \frac{2700}{1350} = 2$.
(c) Equal pulling force $F$: $U = \frac{F^2}{2k} \propto \frac{1}{k} \implies \frac{U_1′}{U_2′} = \frac{k_2}{k_1} = \frac{1350}{2700} = 0.5$.
Example 3 (Equilibrium Stability & Small Oscillations from $U(x)$):
A particle of mass $m = 1.0\text{ kg}$ moves along the $x$-axis in potential $U(x) = x^4 – 8x^2 + 12\text{ J}$.
(a) Find all equilibrium positions.
(b) Classify the stability of each equilibrium position.
(c) Find the frequency $\omega$ of small harmonic oscillations about the stable equilibrium positions.
Solution:
(a) $F = -\frac{dU}{dx} = -(4x^3 – 16x) = 0 \implies 4x(x^2 – 4) = 0 \implies x = 0, \pm 2.0\text{ m}$.
(b) $\frac{d^2U}{dx^2} = 12x^2 – 16$.
– At $x = 0$: $\frac{d^2U}{dx^2} = -16 < 0 \implies$ Local maximum $\implies$ Unstable Equilibrium.
– At $x = \pm 2.0\text{ m}$: $\frac{d^2U}{dx^2} = 12(4) – 16 = +32 > 0 \implies$ Local minimum $\implies$ Stable Equilibrium.
(c) Small oscillation frequency about $x = \pm 2.0\text{ m}$:
$k_{\text{eff}} = \left.\frac{d^2U}{dx^2}\right|_{x=\pm 2} = 32\text{ N/m} \implies \omega = \sqrt{\frac{k_{\text{eff}}}{m}} = \sqrt{\frac{32}{1.0}} = 4\sqrt{2} \approx 5.66\text{ rad/s}$.
Example 4 (Combined Gravitational & Spring Potential Energy):
A block of mass $m = 2.0\text{ kg}$ is dropped from height $h = 0.40\text{ m}$ onto an unstretched vertical spring of stiffness $k = 1960\text{ N/m}$. Taking $g = 9.8\text{ m/s}^2$:
(a) Calculate the maximum spring compression $x_{\text{max}}$.
(b) Find the compression $x_{\text{eq}}$ where the speed is maximum, and find this maximum speed $v_{\text{max}}$.
Solution:
(a) By energy conservation from release to maximum compression ($v_i = 0, v_f = 0$):
$mg(h + x_{\text{max}}) = \frac{1}{2}k x_{\text{max}}^2 \implies 980 x_{\text{max}}^2 – 19.6 x_{\text{max}} – 7.84 = 0 \implies 250 x_{\text{max}}^2 – 5 x_{\text{max}} – 2 = 0$.
$x_{\text{max}} = \frac{5 + \sqrt{25 + 2000}}{500} = \frac{5 + 45}{500} = 0.10\text{ m} = 10.0\text{ cm}$.
(b) Maximum speed occurs when net acceleration is zero ($mg = k x_{\text{eq}}$):
$x_{\text{eq}} = \frac{mg}{k} = \frac{19.6}{1960} = 0.010\text{ m} = 1.0\text{ cm}$.
$mgh = \frac{1}{2}m v_{\text{max}}^2 – mg x_{\text{eq}} + \frac{1}{2}k x_{\text{eq}}^2 \implies 7.84 = v_{\text{max}}^2 – 0.098 \implies v_{\text{max}}^2 = 7.938 \implies v_{\text{max}} \approx 2.82\text{ m/s}$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
The potential energy of a particle in a 1D conservative field is given by $U(x) = 2x^2 – 8x\text{ Joules}$. The position of equilibrium is:
(A) $x = 1.0\text{ m}$
(B) $x = 2.0\text{ m}$
(C) $x = 4.0\text{ m}$
(D) $x = 0$
Problem 2 (JEE Main – Single Correct):
When a spring is stretched by $2.0\text{ cm}$, its stored elastic potential energy is $U$. If the spring is stretched by $8.0\text{ cm}$, the stored potential energy will be:
(A) $4U$
(B) $8U$
(C) $16U$
(D) $2U$
Problem 3 (JEE Main – Single Correct):
A uniform chain of mass $M$ and length $L$ is held on a smooth horizontal table with $\frac{1}{n}$ of its length hanging over the edge. The work done in pulling the hanging part slowly back onto the table is:
(A) $\frac{M g L}{n}$
(B) $\frac{M g L}{2n}$
(C) $\frac{M g L}{2n^2}$
(D) $\frac{M g L}{n^2}$
Problem 4 (JEE Main – Single Correct):
For a system to be in stable equilibrium at position $x_0$, the potential energy $U(x)$ must satisfy:
(A) $\frac{dU}{dx} = 0$ and $\frac{d^2U}{dx^2} > 0$
(B) $\frac{dU}{dx} = 0$ and $\frac{d^2U}{dx^2} < 0$
(C) $\frac{dU}{dx} > 0$ and $\frac{d^2U}{dx^2} = 0$
(D) $\frac{dU}{dx} = 0$ and $\frac{d^2U}{dx^2} = 0$
Problem 5 (JEE Main – Single Correct):
Two springs $A$ and $B$ have spring constants $k_A$ and $k_B$ respectively ($k_A > k_B$). If they are stretched by the same tensile force $F$, which spring stores more potential energy?
(A) Spring $A$
(B) Spring $B$
(C) Both store equal potential energy
(D) Depends on natural lengths of the springs
Problem 6 (JEE Advanced – One or More Correct):
Which of the following statements is/are TRUE regarding potential energy?
(A) Potential energy can be defined only for conservative forces.
(B) The physical force is given by the negative spatial gradient of potential energy: $\vec{F} = -\vec{\nabla}U$.
(C) Changing the reference datum by adding a constant $C$ to $U(\vec{r})$ does not alter the physical force.
(D) The potential energy of an ideal stretched spring can be negative depending on coordinate choice.
Problem 7 (JEE Advanced – One or More Correct):
The potential energy between two atoms in a diatomic molecule as a function of interatomic separation $r$ is given by $U(r) = \frac{A}{r^{12}} – \frac{B}{r^ $U(r) = \frac{A}{r^{12}} – \frac{B}{r^6}$ (where $A, B > 0$). Which of the following is/are correct?
(A) The equilibrium separation is $r_0 = \left(\frac{2A}{B}\right)^{1/6}$.
(B) The equilibrium at $r_0$ is stable.
(C) The binding (dissociation) energy is $D = \frac{B^2}{4A}$.
(D) The interatomic force is attractive for $r > r_0$ and repulsive for $r < r_0$.
Problem 8 (JEE Advanced – One or More Correct):
A block of mass $m$ is released from height $h$ above an uncompressed vertical spring of stiffness $k$. During the subsequent downward motion:
(A) The speed of the block is maximum when it first touches the spring.
(B) The speed of the block is maximum when the spring compression is $x = \frac{mg}{k}$.
(C) The acceleration of the block is zero when its speed is maximum.
(D) The maximum spring compression $x_{\text{max}}$ is strictly greater than $\frac{2mg}{k}$.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A spring has a spring constant of $k = 500.0\text{ N/m}$. The elastic potential energy in Joules stored in the spring when it is compressed by $x = 0.20\text{ m}$ from its natural length is $U$. Find $U$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A particle moves in a 1D conservative field with potential energy $U(x) = 2x^3 – 9x^2 + 12x\text{ Joules}$. Determine the coordinate $x$ in meters corresponding to the position of stable equilibrium.
Solutions & Explanations
Answer Key Summary:
1. (B) | 2. (C) | 3. (C) | 4. (A) | 5. (B) | 6. (A, B, C) | 7. (A, B, C, D) | 8. (B, C, D) | 9. 10 | 10. 2
Solution 1:
$F = -\frac{dU}{dx} = -(4x – 8) = 0 \implies x = 2.0\text{ m}$. Correct: (B).
Solution 2:
$U \propto x^2 \implies U_2 = \left(\frac{8}{2}\right)^2 U = 16U$. Correct: (C).
Solution 3:
$W = \frac{M}{n} g \left(\frac{L}{2n}\right) = \frac{MgL}{2n^2}$. Correct: (C).
Solution 4:
Stable equilibrium is at local potential energy minimum ($\frac{dU}{dx} = 0, \frac{d^2U}{dx^2} > 0$). Correct: (A).
Solution 5:
$U = \frac{F^2}{2k} \propto \frac{1}{k}$. Since $k_B < k_A$, spring $B$ stores more energy. Correct: (B).
Solution 6:
(A), (B), and (C) are verified fundamental properties. (D) is false because $U_s = \frac{1}{2}kx^2 \ge 0$ is positive definite. Correct: (A, B, C).
Solution 7:
All statements (A, B, C, D) describe the complete calculus and physical properties of the Lennard-Jones molecular potential. Correct: (A, B, C, D).
Solution 8:
(A) False: downward acceleration continues until upward spring force matches gravity. (B), (C), and (D) are verified dynamical properties. Correct: (B, C, D).
Solution 9:
$U = \frac{1}{2}kx^2 = \frac{1}{2}(500)(0.20^2) = 10.0\text{ Joules}$. Correct: 10.
Solution 10:
$\frac{dU}{dx} = 6(x-1)(x-2) = 0 \implies x = 1, 2$. $\frac{d^2U}{dx^2} = 12x – 18$. At $x = 2$, $\frac{d^2U}{dx^2} = +6 > 0$ (stable). Correct: 2.