Work and Potential Energy; Conservation

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2017)
• Priority tier: T4
• Marks (count): 16 (1)
• Average solve time: ~24 min
• Difficulty mix: medium 1
• Section: B | Dominant type: derivation

Why This Chapter Matters

This atom appeared once (2017) as a Section B question worth 16 marks — a full energy-methods derivation. UPSC uses it to test whether you can connect the line-integral definition of work to the potential-energy function, derive the work–energy theorem, and apply conservation of energy to a concrete mechanical system. A structured derivation with a numerical application earns full marks.

Minimum Theory

Work done by a force

For a particle moving along a path $C$, the work done by force $\mathbf{F}$ is:

$W = \int_C \mathbf{F} \cdot d\mathbf{r} = \int_C (F_x\,dx + F_y\,dy + F_z\,dz)$

For a constant force: $W = \mathbf{F}\cdot\Delta\mathbf{r} = F\,d\cos\alpha$, where $\alpha$ is the angle between $\mathbf{F}$ and displacement.

Kinetic energy and work–energy theorem

Kinetic energy: $T = \tfrac{1}{2}m|\mathbf{v}|^2$.

From Newton’s second law $\mathbf{F} = m\ddot{\mathbf{r}}$:

$W_{A\to B} = \int_A^B \mathbf{F}\cdot d\mathbf{r} = \int_A^B m\ddot{\mathbf{r}}\cdot\dot{\mathbf{r}}\,dt = \frac{d}{dt}\left(\tfrac{1}{2}m|\mathbf{v}|^2\right)\,dt = T_B - T_A$

This is the work–energy theorem: work done equals change in kinetic energy.

Conservative forces and potential energy

A force $\mathbf{F}$ is conservative if $\oint_C \mathbf{F}\cdot d\mathbf{r} = 0$ for every closed path, equivalently $\nabla\times\mathbf{F} = \mathbf{0}$.

For a conservative force, there exists a scalar potential energy $V(\mathbf{r})$ such that:

$\mathbf{F} = -\nabla V, \quad \text{i.e.,}\quad F_x = -\frac{\partial V}{\partial x},\; F_y = -\frac{\partial V}{\partial y},\; F_z = -\frac{\partial V}{\partial z}$

Work done by a conservative force:

$W_{A\to B} = \int_A^B \mathbf{F}\cdot d\mathbf{r} = -\int_A^B dV = V_A - V_B$

Conservation of mechanical energy

Combining the work–energy theorem with the potential energy expression (all forces conservative):

$T_B - T_A = V_A - V_B \implies T_A + V_A = T_B + V_B$

$\boxed{E = T + V = \text{constant}}$

Common potential energies

• Gravity (near Earth): $V = mgh$ (h above datum)
• Elastic spring: $V = \tfrac{1}{2}kx^2$ (x = extension from natural length)
• Gravitational (point mass): $V = -\dfrac{GMm}{r}$

Power

$P = \frac{dW}{dt} = \mathbf{F}\cdot\mathbf{v}$

Units: watts (W) = J/s.

Question Archetypes

energy-conservation-derivation (1 question; 2017)

Recognition Cues

• The question involves a particle on an incline, attached to a spring, or moving under gravity.
• You are asked to “prove” or “derive” the energy equation, then apply it.
• The phrase “potential energy” or “conservative force” appears.
• A specific mechanical quantity (speed, maximum compression, height reached) is to be found.

Solution Template

1. Identify all forces; determine which are conservative.
2. Write $T = \tfrac{1}{2}mv^2$ and the appropriate $V$ for each conservative force.
3. Apply $T + V = E$ at two key positions (usually initial state and state of interest).
4. Solve the resulting algebraic equation for the unknown.
5. State the answer with correct units; verify dimensions.

Worked Example

2017 Paper 1, 2017-P1-Q8b (16 marks)

A particle of mass $m$ is attached to one end of a light elastic spring of natural length $l$ and modulus of elasticity $\lambda$. The other end of the spring is fixed to a point $A$ on a smooth horizontal table. The particle is held at rest at a distance $l + a$ from $A$ and released. Derive the equation of motion from energy principles and find the speed of the particle when it returns to the natural length position.

Step 1. Set up coordinates and forces.

Let $x$ denote the extension of the spring beyond natural length $l$. At initial release, $x = a$; at natural length, $x = 0$.

The elastic restoring force is $F = -\dfrac{\lambda}{l}\,x$ (Hooke’s law, tension positive towards $A$).

This is conservative with potential energy:

$V = \frac{\lambda x^2}{2l}$

(since $F = -dV/dx$). Kinetic energy: $T = \tfrac{1}{2}m\dot{x}^2$.

Step 2. State total energy at initial position.

At $x = a$, $\dot{x} = 0$:

$E = T + V = 0 + \frac{\lambda a^2}{2l} = \frac{\lambda a^2}{2l}$

Step 3. Apply conservation of energy.

At general extension $x$:

$\frac{1}{2}m\dot{x}^2 + \frac{\lambda x^2}{2l} = \frac{\lambda a^2}{2l}$

Differentiating with respect to $t$ gives the equation of motion:

$m\ddot{x} + \frac{\lambda}{l}x = 0 \implies \ddot{x} + \omega^2 x = 0, \quad \omega^2 = \frac{\lambda}{ml}$

This is simple harmonic motion.

Step 4. Speed at natural length ($x = 0$).

Setting $x = 0$ in the energy equation:

$\frac{1}{2}mv^2 + 0 = \frac{\lambda a^2}{2l} \implies v^2 = \frac{\lambda a^2}{ml}$

$\boxed{v = a\sqrt{\frac{\lambda}{ml}}}$

Common Traps

• Using $V = \tfrac{1}{2}kx^2$ with $k$ without checking whether the problem gives $k$ directly or $\lambda/l$ (modulus form); confusing the two conventions.
• Forgetting to set $T = 0$ at the initial rest position, leading to an incorrect value of $E$.
• Taking $V = 0$ at the natural length position and then failing to account for the potential at other positions consistently.
• Differentiating the energy equation to get the equation of motion but making a sign error on the restoring-force term.

Marks-Aware Writing

This is a 16-mark Section B derivation. Examiners expect:

1. Definition of $T$ and $V$ with correct formulas for the specific system (3 marks).
2. Calculation of total energy $E$ from initial conditions (2 marks).
3. Conservation equation written in full at general position (3 marks).
4. Equation of motion derived by differentiation (3 marks).
5. Application — substituting the specific position and solving for the unknown (3 marks).
6. Correct final answer with units (2 marks).

Show every algebraic step; do not jump from the energy equation to the answer without showing the substitution.

Practice Set

Only one historical question on this atom (shown above).