Concept Card: Conservative vs. Non-Conservative Forces & Mechanical Energy Conservation
1. Conservative Forces – Definitions & Criteria:
A force field $\vec{F}_c$ is conservative if the work done by it in moving a particle between two points depends solely on the initial and final position coordinates, being completely independent of the path taken.
- Closed-Loop Criterion: The line integral around any arbitrary closed loop is zero:
$$\oint \vec{F}_c \cdot d\vec{r} = 0$$ - Differential (Curl) Criterion in 3D: $\vec{\nabla} \times \vec{F}_c = \vec{0}$ everywhere:
$\frac{\partial F_z}{\partial y} = \frac{\partial F_y}{\partial z}, \quad \frac{\partial F_x}{\partial z} = \frac{\partial F_z}{\partial x}, \quad \frac{\partial F_y}{\partial x} = \frac{\partial F_x}{\partial y}$ - Scalar Potential Energy Relation:
$\vec{F}_c = -\vec{\nabla}U \iff W_c = -\Delta U = U_i – U_f$ - Examples: Gravitational force, electrostatic Coulomb force, ideal spring restoring force, any central force $\vec{F}(r) = f(r)\hat{r}$.
2. Non-Conservative Forces:
A force is non-conservative if the work done by it depends on the actual trajectory followed:
$$\oint \vec{F}_{nc} \cdot d\vec{r} \neq 0$$
- Cannot be expressed as the negative gradient of a potential energy function ($U$ does not exist).
- Dissipative forces convert macroscopic mechanical energy into microscopic thermal energy, sound, or internal deformation.
- Examples: Kinetic friction ($W_f = -\int f_k\,ds < 0$), viscous fluid drag, air resistance.
3. The Principle of Conservation of Mechanical Energy:
Total mechanical energy is defined as $E = K + U$.
From the Work-Energy Theorem ($W_c + W_{nc} + W_{\text{ext}} = \Delta K$ with $W_c = -\Delta U$):
$$W_{nc} + W_{\text{ext}} = \Delta K + \Delta U = \Delta E = E_f – E_i$$
- Isolated Conservative Systems ($W_{nc} = 0, W_{\text{ext}} = 0$):
$\Delta E = 0 \implies E_i = E_f \implies K_i + U_i = K_f + U_f = \text{constant}$
$$\frac{1}{2}m v_i^2 + U(\vec{r}_i) = \frac{1}{2}m v_f^2 + U(\vec{r}_f)$$ - Dissipative Systems ($W_{nc} = W_f < 0$):
$E_f – E_i = W_f \implies E_f = E_i – |W_f|$
(Loss in mechanical energy equals the work done against friction).
4. Potential Energy Curves & Turning Points:
For a particle of total mechanical energy $E$ moving in potential $U(x)$:
- Allowed Region: $E \ge U(x)$, ensuring kinetic energy $K(x) = E – U(x) \ge 0$.
- Turning Points: Points where $E = U(x) \implies K = 0 \implies v = 0$. The particle momentarily halts and reverses its direction of motion.
- Forbidden Region: Domains where $E < U(x)$ (classically inaccessible).
- Speed at any point: $v(x) = \sqrt{\frac{2(E – U(x))}{m}}$.
Solved Examples
Example 1 (Testing Conservativeness & Path Dependence):
Consider two 2D force fields:
$\vec{F}_1 = (3x^2 y)\hat{i} + (x^3)\hat{j}\text{ N}$
$\vec{F}_2 = (3x^2 y)\hat{i} + (y^3)\hat{j}\text{ N}$
(a) Determine which of these forces is conservative and which is non-conservative.
(b) For the conservative force, derive its potential function $U(x, y)$ taking $U(0, 0) = 0$, and find the work done moving from $(0, 0)$ to $(2, 3)$.
(c) For the non-conservative force, calculate work done moving from $(0, 0)$ to $(1, 1)$ along Path $A$ ($y = x$) and Path $B$ ($y = x^2$).
Solution:
(a) Checking $\frac{\partial F_y}{\partial x} – \frac{\partial F_x}{\partial y}$:
– For $\vec{F}_1$: $\frac{\partial F_{1y}}{\partial x} = 3x^2, \quad \frac{\partial F_{1x}}{\partial y} = 3x^2 \implies \vec{\nabla}\times\vec{F}_1 = \vec{0}$ (Conservative).
– For $\vec{F}_2$: $\frac{\partial F_{2y}}{\partial x} = 0, \quad \frac{\partial F_{2x}}{\partial y} = 3x^2 \neq 0 \implies \vec{\nabla}\times\vec{F}_2 \neq \vec{0}$ (Non-conservative).
(b) For $\vec{F}_1$: $-\frac{\partial U}{\partial x} = 3x^2 y \implies U(x, y) = -x^3 y + g(y) \implies -\frac{\partial U}{\partial y} = x^3 \implies U(x, y) = -x^3 y$.
Work done: $W_1 = U(0, 0) – U(2, 3) = 0 – [-(2^3)(3)] = +24.0\text{ Joules}$.
(c) For $\vec{F}_2$:
– Along Path $A$ ($y = x \implies dy = dx$): $W_A = \int_0^1 (3x^3 + x^3)\,dx = \int_0^1 4x^3\,dx = [x^4]_0^1 = 1.0\text{ J}$.
– Along Path $B$ ($y = x^2 \implies dy = 2x\,dx$): $W_B = \int_0^1 [3x^4 + 2x^7]\,dx = \left[\frac{3}{5}x^5 + \frac{2}{8}x^8\right]_0^1 = 0.60 + 0.25 = 0.85\text{ J}$.
$W_A \neq W_B$, confirming path dependence.
Example 2 (Mechanical Energy Conservation & Incline Launch):
A block of mass $m = 2.0\text{ kg}$ is compressed against a horizontal spring of constant $k = 800\text{ N/m}$ by $x_0 = 0.20\text{ m}$ on a smooth floor and released from rest. It slides up a smooth incline of $\theta = 30^\circ$. Taking $g = 10\text{ m/s}^2$:
(a) Find the speed $v_0$ as the block leaves the spring.
(b) Find the maximum vertical height $h$ and distance $s$ along the incline.
(c) Find the distance $s’$ if the incline has friction $\mu_k = 0.20$.
Solution:
(a) Initial elastic energy: $E_i = \frac{1}{2}kx_0^2 = \frac{1}{2}(800)(0.20^2) = 16.0\text{ J}$.
At spring release: $\frac{1}{2}mv_0^2 = 16.0 \implies \frac{1}{2}(2.0)v_0^2 = 16.0 \implies v_0 = 4.0\text{ m/s}$.
(b) On smooth incline: $mgh = E_i \implies 2.0(10)h = 16.0 \implies h = 0.80\text{ m}$.
Distance along incline: $s = \frac{h}{\sin 30^\circ} = \frac{0.80}{0.50} = 1.60\text{ m}$.
(c) With friction on incline: $W_f = -\mu_k mg\cos(30^\circ) s’ = -0.20(20)\left(\frac{\sqrt{3}}{2}\right) s’ = -2\sqrt{3} s’$.
$mgs’\sin 30^\circ – 16.0 = -2\sqrt{3}s’ \implies (10 + 2\sqrt{3})s’ = 16.0 \implies s’ = \frac{16.0}{13.464} \approx 1.188\text{ m}$.
Example 3 (Potential Energy Well, Turning Points & Bound Motion):
A particle of mass $m = 1.0\text{ kg}$ moves along the $x$-axis in potential $U(x) = \frac{a}{x^2} – \frac{b}{x}$ ($x > 0$), with $a = 2.0\text{ J}\cdot\text{m}^2$ and $b = 4.0\text{ J}\cdot\text{m}$.
(a) Find the equilibrium position $x_0$ and minimum potential energy $U_{\text{min}}$.
(b) For total energy $E = -1.0\text{ J}$, find the turning points $x_1$ and $x_2$.
(c) Find the maximum kinetic energy and maximum speed.
Solution:
(a) $F = -\frac{dU}{dx} = \frac{2a}{x^3} – \frac{b}{x^2} = 0 \implies x_0 = \frac{2a}{b} = \frac{2(2.0)}{4.0} = 1.0\text{ m}$.
$U_{\text{min}} = U(1.0) = \frac{2.0}{1.0} – \frac{4.0}{1.0} = -2.0\text{ Joules}$.
(b) Turning points occur where $U(x) = E \implies \frac{2.0}{x^2} – \frac{4.0}{x} = -1.0 \implies x^2 – 4x + 2 = 0$.
$x = \frac{4 \pm \sqrt{16 – 8}}{2} = 2 \pm \sqrt{2} \implies x_1 = 2 – \sqrt{2} \approx 0.586\text{ m}, \quad x_2 = 2 + \sqrt{2} \approx 3.414\text{ m}$.
(c) $K_{\text{max}} = E – U_{\text{min}} = -1.0 – (-2.0) = +1.0\text{ Joule}$.
$v_{\text{max}} = \sqrt{\frac{2K_{\text{max}}}{m}} = \sqrt{\frac{2(1.0)}{1.0}} = \sqrt{2} \approx 1.414\text{ m/s}$.
Example 4 (Spring-Block System with Frictional Dissipation):
A block of mass $m = 4.0\text{ kg}$ is pressed against a horizontal spring of stiffness $k = 400\text{ N/m}$, compressing it by $x_0 = 0.50\text{ m}$ on a rough horizontal floor with $\mu_k = 0.20$. Released from rest, taking $g = 10\text{ m/s}^2$:
(a) Find the speed of the block as it passes through the unstretched position ($x = 0$).
(b) If attached to the spring, find the maximum elongation $x_1$ on the return stroke.
(c) If unattached, find the total distance $D$ traveled before stopping permanently.
Solution:
(a) $E_i = \frac{1}{2}kx_0^2 = 50.0\text{ J}$. Friction work to $x = 0$: $W_f = -\mu_k mg x_0 = -0.20(40)(0.50) = -4.0\text{ J}$.
$E_0 = 50.0 – 4.0 = 46.0\text{ J} \implies \frac{1}{2}(4.0)v_0^2 = 46.0 \implies v_0 = \sqrt{23.0} \approx 4.80\text{ m/s}$.
(b) If attached: $\Delta E = W_f \implies 200x_1^2 – 50.0 = -4.0 – 8.0x_1 \implies 100x_1^2 + 4.0x_1 – 23.0 = 0$.
$x_1 = \frac{-4.0 + \sqrt{16 + 9200}}{200} = \frac{-4.0 + 96.0}{200} = 0.46\text{ m} = 46.0\text{ cm}$.
(c) If unattached: Slides freely after $x = 0$ with $K_0 = 46.0\text{ J}$. Additional distance $s = \frac{K_0}{\mu_k mg} = \frac{46.0}{8.0} = 5.75\text{ m}$.
Total distance: $D = x_0 + s = 0.50 + 5.75 = 6.25\text{ meters}$.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A force $\vec{F}$ is conservative if:
(A) The work done by $\vec{F}$ around any closed loop is zero ($\oint \vec{F}\cdot d\vec{r} = 0$)
(B) The force depends explicitly on the velocity of the particle
(C) The work done is directly proportional to the path length
(D) It always opposes the motion of the body
Problem 2 (JEE Main – Single Correct):
A pendulum bob of mass $m$ is released from rest from a horizontal position ($\theta = 90^\circ$ to the vertical). If the string has length $L$, the speed of the bob at the lowest point is:
(A) $\sqrt{g L}$
(B) $\sqrt{2 g L}$
(C) $2\sqrt{g L}$
(D) $\sqrt{\frac{g L}{2}}$
Problem 3 (JEE Main – Single Correct):
When a non-conservative force (such as friction) acts on a system, the change in total mechanical energy $\Delta E = \Delta(K + U)$ is:
(A) Always strictly positive
(B) Equal to the work done by the non-conservative force ($W_{nc}$)
(C) Always identically zero
(D) Equal to the total potential energy of the system
Problem 4 (JEE Main – Single Correct):
A block of mass $m$ slides down a rough incline. During the motion, the change in kinetic energy is $\Delta K = +40\text{ J}$ and the change in gravitational potential energy is $\Delta U = -100\text{ J}$. The work done by the frictional force is:
(A) $+60\text{ J}$
(B) $-60\text{ J}$
(C) $-140\text{ J}$
(D) $+140\text{ J}$
Problem 5 (JEE Main – Single Correct):
At a turning point of a particle’s motion in a 1D conservative potential $U(x)$:
(A) Kinetic energy is maximum
(B) Potential energy is zero
(C) Total mechanical energy equals potential energy ($E = U(x)$)
(D) Acceleration is zero
Problem 6 (JEE Advanced – One or More Correct):
Which of the following conditions is/are necessary and sufficient for a force field $\vec{F}$ to be conservative?
(A) The work done between any two points is independent of the path.
(B) $\vec{\nabla} \times \vec{F} = \vec{0}$ everywhere in the domain.
(C) The force can be written as $\vec{F} = -\vec{\nabla}U$ for some scalar function $U$.
(D) The magnitude of the force must be constant everywhere.
Problem 7 (JEE Advanced – One or More Correct):
A particle of mass $m$ moves in a 1D potential well $U(x) = \frac{1}{2}kx^2$ with total energy $E$:
(A) The motion is bounded between turning points $x = \pm\sqrt{\frac{2E}{k}}$.
(B) The kinetic energy is maximum at $x = 0$ with value $K_{\text{max}} = E$.
(C) The acceleration at the turning points is non-zero.
(D) The mechanical energy $E$ is conserved throughout the motion.
Problem 8 (JEE Advanced – One or More Correct):
A block attached to an ideal spring oscillates on a rough horizontal surface:
(A) Mechanical energy is continually dissipated into thermal energy.
(B) The amplitude of oscillation decreases with time.
(C) The Work-Energy Theorem ($W_{\text{net}} = \Delta K$) is invalid because friction is non-conservative.
(D) The total energy of the universe (mechanical + thermal) is strictly conserved.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A block of mass $m = 2.0\text{ kg}$ is dropped from a height $h = 0.80\text{ m}$ above an uncompressed vertical spring of constant $k = 200.0\text{ N/m}$. Taking $g = 10\text{ m/s}^2$, find the maximum compression $x_{\text{max}}$ of the spring in meters.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A conservative 2D force field has potential energy $U(x, y) = 3x^2 y – y^3\text{ Joules}$. The magnitude of the force in Newtons at the point $(x, y) = (1.0\text{ m}, 2.0\text{ m})$ is $F$. Calculate $F^2$.
Solutions & Explanations
Answer Key Summary:
1. (A) | 2. (B) | 3. (B) | 4. (B) | 5. (C) | 6. (A, B, C) | 7. (A, B, C, D) | 8. (A, B, D) | 9. 0.8 | 10. 225
Solution 1:
A force is conservative if and only if its closed-loop line integral vanishes: $\oint \vec{F}\cdot d\vec{r} = 0$. Correct: (A).
Solution 2:
$mgh = \frac{1}{2}mv^2 \implies v = \sqrt{2gh} = \sqrt{2gL}$. Correct: (B).
Solution 3:
$W_c + W_{nc} = \Delta K \implies -\Delta U + W_{nc} = \Delta K \implies W_{nc} = \Delta K + \Delta U = \Delta E$. Correct: (B).
Solution 4:
$W_f = \Delta E = \Delta K + \Delta U = (+40\text{ J}) + (-100\text{ J}) = -60\text{ Joules}$. Correct: (B).
Solution 5:
At turning points, the particle halts momentarily: $v = 0 \implies K = 0 \implies E = U(x)$. Correct: (C).
Solution 6:
(A), (B), and (C) are equivalent criteria defining conservative forces. Magnitude need not be constant (e.g., spring force $-kx$). Correct: (A, B, C).
Solution 7:
All statements (A, B, C, D) are verified mathematical and dynamical properties of a harmonic potential well. Correct: (A, B, C, D).
Solution 8:
(A), (B), and (D) are true. (C) is false because WET holds universally for all forces. Correct: (A, B, D).
Solution 9:
$mg(h + x) = \frac{1}{2}kx^2 \implies 20(0.80 + x) = 100x^2 \implies 25x^2 – 5x – 4 = 0 \implies (5x – 4)(5x + 1) = 0 \implies x = 0.80\text{ m}$. Correct: 0.8.
Solution 10:
$F_x = -\frac{\partial U}{\partial x} = -6xy = -12\text{ N}$.
$F_y = -\frac{\partial U}{\partial y} = 3y^2 – 3x^2 = 3(4) – 3(1) = +9\text{ N}$.
$F^2 = (-12)^2 + 9^2 = 144 + 81 = 225\text{ N}^2$. Correct: 225.