Work-energy theorem

At a Glance

• Frequency: 2 sub-parts across 2 of 13 years (2017, 2019)
• Priority tier: T3
• Marks (count): 10 (2)
• Average solve time: ~8 min
• Difficulty mix: easy 1, medium 1
• Section: B | Dominant type: computation

Why This Chapter Matters

Both UPSC questions on this atom are compulsory 10-mark items from Section B — fast, reliable marks. The 2019 question is a straightforward work-as-integral problem that takes under six minutes. The 2017 question uses energy conservation to derive a motion equation on a cardioid, testing whether you can correctly identify gravitational depth and elastic potential energy. Together they form two reusable templates that cover every work-energy question the examiner has written.

Minimum Theory

Work-energy theorem. The work done by a force $\mathbf{F}$ as a particle moves from $A$ to $B$ along a path equals the change in kinetic energy:

$W_{A\to B} = T_B - T_A = \int_A^B \mathbf{F}\cdot d\mathbf{r}.$

For a conservative force with potential $V$ (where $\mathbf{F} = -\nabla V$), this is simply $W = V_A - V_B$.

Conservation of energy. If all forces are conservative (gravity, elastic string, etc.), the total mechanical energy is constant: $T + V = E$. Set $E$ using the initial conditions at the release point; then write $T + V_\text{grav} + V_\text{elastic} = E$ at a general position to get the equation of motion.

Inverse-square gravity. At distance $r$ from the centre of the earth, the attractive force per unit mass is $\mu/r^2$ (or, for a particle of weight $W$ on the surface of a body of radius $h$: calibrate $k = Wh^2$ so the force at distance $r$ is $k/r^2$). The work done in falling from $r = r_1$ to $r = r_2 < r_1$ is

$W = k\!\left(\frac{1}{r_2} - \frac{1}{r_1}\right).$

Elastic potential energy. For a string of natural length $a$ and modulus of elasticity $\lambda$, the elastic PE of extension $e$ is $\lambda e^2/(2a)$.

Question Archetypes

energy-conservation (1 question(s); 2017)

Recognition Cues

• Ring or particle on a named curve (cardioid, cycloid, etc.) with a spring/elastic string attached.
• “Released from rest when the string is [horizontal/vertical]”; find the energy equation at a general position.
• The question says “show by energy conservation that $f(\theta,\dot\theta) = 0$.”

Solution Template

1. Write the speed $v^2 = \dot r^2 + r^2\dot\theta^2$ in terms of $\dot\theta$ using the curve equation $r = r(\theta)$.
2. Identify the gravitational PE using the depth below a fixed datum (downward vertical = direction of positive depth).
3. Compute the elastic PE: extension $e = r - a_0$ (natural length $a_0$); PE $= \lambda e^2/(2a_0)$.
4. Write $T + V_\text{grav} + V_\text{elastic} = 0$ (since total energy is zero: released from rest at a configuration with zero PE).
5. Divide by common factors to obtain the stated equation.

Worked Example

2017 Paper 1, 2017-P1-Q5c (10 marks)

A fixed wire is in the shape of the cardioid $r=a(1+\cos\theta)$, the initial line being the downward vertical. A small ring of mass $m$ slides on the wire and is attached to the point $r=0$ of the cardioid by an elastic string of natural length $a$ and modulus of elasticity $4mg$. The string is released from rest when the string is horizontal. Show by energy conservation that $a\dot\theta^2(1+\cos\theta)-g\cos\theta(1-\cos\theta)=0$.

Step 1 — Speed. With $r = a(1+\cos\theta)$, $\dot r = -a\sin\theta\,\dot\theta$:

$v^2 = \dot r^2 + r^2\dot\theta^2 = a^2\sin^2\theta\,\dot\theta^2 + a^2(1+\cos\theta)^2\dot\theta^2 = 2a^2\dot\theta^2(1+\cos\theta).$

$T = \tfrac12 m v^2 = ma^2\dot\theta^2(1+\cos\theta).$

Step 2 — Gravitational PE. The initial line is the downward vertical, so the depth below $O$ is $r\cos\theta = a(1+\cos\theta)\cos\theta$. Taking $O$ as datum:

$V_g = -mg\cdot a(1+\cos\theta)\cos\theta.$

Step 3 — Elastic PE. Natural length $a$, modulus $\lambda=4mg$, current length $r=a(1+\cos\theta)$, extension $e = a\cos\theta$:

$V_e = \frac{4mg}{2a}(a\cos\theta)^2 = 2mga\cos^2\theta.$

Step 4 — Zero initial energy. At release, the string is horizontal, so the radius vector is horizontal. Since the initial line is the downward vertical, horizontal means $\theta = \pi/2$. There $r = a$, $e = 0$, $T = 0$, $V_g = 0$, $V_e = 0$, hence $E_0 = 0$.

Step 5 — Conservation. $T + V_g + V_e = 0$:

$ma^2\dot\theta^2(1+\cos\theta) - mga(1+\cos\theta)\cos\theta + 2mga\cos^2\theta = 0.$

Divide by $ma$:

$a\dot\theta^2(1+\cos\theta) - g\cos\theta(1+\cos\theta - 2\cos\theta) = 0.$

$\boxed{a\dot\theta^2(1+\cos\theta) - g\cos\theta(1-\cos\theta) = 0.}$

Common Traps

• “String horizontal” means the radius vector is horizontal, i.e., $\theta = \pi/2$ (the angle is measured from the downward vertical). Setting $\theta = 0$ gives the wrong release point.
• Depth for PE must be measured along the downward vertical: the $r\cos\theta$ component gives the depth. A sign error here flips the whole result.
• Elastic PE uses $\lambda e^2/(2a)$ with $e = r - a = a\cos\theta$; the specific modulus $4mg$ is tuned to produce the clean cancellation into $(1-\cos\theta)$.
• Verify the zero-energy claim: at $\theta = \pi/2$, all three energy terms vanish simultaneously — this is the self-check.

work-done-integral (1 question(s); 2019)

Recognition Cues

• “Height $h$ above the surface”; “falls to the surface under inverse-square attraction.”
• Surface weight $W$ given; earth’s radius also given (often the same letter $h$).
• “Show the work done is [a specific expression].”

Solution Template

1. Write the force law $F(r) = k/r^2$.
2. Calibrate $k$ from the surface weight: $W = k/h^2 \Rightarrow k = Wh^2$.
3. Convert “height above surface” to “distance from centre”: add the earth’s radius.
4. Integrate $W = \int_{r_\text{end}}^{r_\text{start}} k/r^2\,dr = k[-1/r]$.
5. Substitute limits and simplify.

Worked Example

2019 Paper 1, 2019-P1-Q5d (10 marks)

The force of attraction of a particle by the earth is inversely proportional to the square of its distance from the earth’s centre. A particle, whose weight on the surface of the earth is $W$, falls to the surface of the earth from a height $3h$ above it. Show that the magnitude of work done by the earth’s attraction force is $\dfrac{3}{4}hW$, where $h$ is the radius of the earth.

Step 1 — Force law and calibration. Force at distance $r$ is $F(r) = k/r^2$. At the surface ($r = h$), $F = W$, so $k = Wh^2$.

Step 2 — Distance from centre. Height $3h$ above the surface means starting distance $r_\text{start} = h + 3h = 4h$ from the centre.

Step 3 — Work integral. Force and displacement both point inward (work is positive):

$\mathcal{W} = \int_h^{4h} \frac{Wh^2}{r^2}\,dr = Wh^2\!\left[-\frac{1}{r}\right]_h^{4h} = Wh^2\!\left(\frac{1}{h} - \frac{1}{4h}\right) = Wh\cdot\frac{3}{4}.$

$\boxed{\mathcal{W} = \frac{3}{4}\,hW.}$

Common Traps

• “Height $3h$ above the surface” gives starting distance $4h$ from the centre, not $3h$. The earth’s radius $h$ must be added.
• All distances appear in the denominator via the $1/r^2$ law — measure from the centre, not the surface.
• The force does positive work on the falling particle (force and motion both inward). The question asks for magnitude, which is positive.

Marks-Aware Writing

A 10-mark answer must: write the correct energy form or force law; set up the computation cleanly; execute the integration (or energy balance) in full; state the final result. In the work-done question, showing the integral and evaluating both limits earns most marks — the setup and result together are worth about 8 marks. In the energy-conservation question, the kinetic energy calculation (Step 1) typically earns 3–4 marks; potential energy terms another 3–4; the final algebraic simplification 2–3.

Practice Set

• 2024-P1-Q7b (15 m) — Energy methods in a dynamics problem. ()
• 2024-P1-Q8c (20 m) — Extended energy/work calculation. ()
• 2013-P1-Q7a (20 m) — Dynamics with energy conservation. ()
• 2014-P1-Q7b (15 m) — Work-energy theorem application. ()
• 2015-P1-Q6d (13 m) — Time to centre under inverse-distance force; energy first integral. ()
• 2021-P1-Q7c (15 m) — Energy conservation in motion. ()
• 2020-P1-Q5e (10 m) — Quick energy calculation. ()
• 2020-P1-Q8c (15 m) — Work done against a force. ()
• 2025-P1-Q6b (15 m) — Variable force energy problem. ()
• 2025-P1-Q8c (20 m) — Extended energy methods. ()