Work, Power & Conservation of Energy: Master Notes

Chapter Overview

Work and Energy provide scalar tools to solve intricate dynamics problems where direct integration of Newton's laws is intractable. In IIT-JEE (Advanced), the primary focus areas are conservative force gradient analysis ($\overrightarrow{F}=-\mathrm{\nabla }U$), vertical circular motion thresholds, and variable spring/chain systems.

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1. Work-Energy Theorem & Power

1.1 The Master Work-Energy Theorem

$$\boxed{{\displaystyle W_{\text{all forces}}=W_{\text{conservative}}+W_{\text{non-conservative}}+W_{\text{pseudo}}=\mathrm{\Delta }K=K_f-K_i}}$$

• Work by a Variable Force in 3D:$$W={\int }_{{\overrightarrow{r}}_1}^{{\overrightarrow{r}}_2}\overrightarrow{F}\cdot d\overrightarrow{r}=\int (F_{x}dx+F_{y}dy+F_{z}dz)$$
• Instantaneous Power:$$P=\frac{dW}{dt}=\overrightarrow{F}\cdot \overrightarrow{v}$$

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2. Conservative Forces & Potential Energy Surfaces $U(x,y,z)$

A force is conservative if and only if the work done around any closed loop is zero: $\oint \overrightarrow{F}\cdot d\overrightarrow{r}=0⟺\mathrm{\nabla }\times \overrightarrow{F}=\overrightarrow{0}$.

$$\mathrm{\Delta }U=-W_{\text{conservative}}=-{\int }_{{\overrightarrow{r}}_1}^{{\overrightarrow{r}}_2}\overrightarrow{F}\cdot d\overrightarrow{r}$$$$\boxed{{\displaystyle \overrightarrow{F}=-\mathrm{\nabla }U=-(\frac{\partial U}{\partial x}\hat{i}+\frac{\partial U}{\partial y}\hat{j}+\frac{\partial U}{\partial z}\hat{k})}}$$

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2.1 Potential Energy Curves $U(x)$ & Stability Analysis

Multi-Mode DiagramWork, Energy & Potential Energy Wells U(x)

Option 1: Publication-Grade Scientific Vector SVG

Potential well U(x) displaying stable, unstable, and neutral equilibrium points alongside turning points E = K + U.

Conservative Force: $F(x) = -\frac{dU}{dx}$

Stable Equilibrium: $\frac{dU}{dx} = 0, \quad \frac{d^2U}{dx^2} > 0$

Option 2: Interactive 60 FPS Canvas Physics Engine

Particle oscillating in potential well exchanging kinetic and potential energy.

Total Energy E: 60 J

Total Energy ($E$):60 J

Option 3: Interactive 3D Spatial Viewport

3D potential surface.

+X

+Y

+Z

👆 Click and drag anywhere to rotate in 3D space

Azimuth Angle (ψ):35°

Elevation Angle (θ):-22°

Conservation of Energy:$E = K + U$

Option 5: Reactive Parametric Physics with Live Calculus

Harmonic oscillator energy calculus.

Spring constant k: 40 N/m

Amplitude A: 5 m

Max Energy ($E_{\max}$):$500.0 J$

Active Mode:Option 1 (Print & Derivation Vector SVG)Switch modes using the tabs above to explore different visualizations.

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3. Vertical Circular Motion: Complete Critical Values

Consider a mass $m$ tied to a light string of length $R$ moving in a vertical circle.

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Figure Vertical Circular Motion: Critical speed thresholds and tension relations for string vs light rod.

Complete Mastery Rules:

1. Critical speed at bottom for complete circle:$$v_{\text{bottom}}\ge \sqrt{5gR}⟹v_{\text{top}}\ge \sqrt{gR}$$
2. Tension at bottom and top:$$T_{\text{bottom}}=\frac{m{v}_{\text{bottom}}^2}{R}+mg\ge 6mg$$$$T_{\text{bottom}}-T_{\text{top}}=6mg\text{(Always invariant for any full loop!)}$$
3. If $\sqrt{2gR}<v_0<\sqrt{5gR}$:
   • The string slacks ($T=0$) in the upper hemisphere before reaching the top at angle $\theta$ from vertical ($\cos \theta =\frac{v_0^2-2gR}{3gR}$).
   • After slacking, the particle executes standard projectile motion under gravity alone!
4. Light Rigid Rod vs Flexible String:
   • For a rigid rod, velocity can reach zero at the top without collapsing:$$v_{\text{bottom, min}}=\sqrt{4gR}=2\sqrt{gR}$$

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4. Authentic Previous Years Questions (PYQs)

PYQ 1: JEE Advanced 2023 (Paper 1) — Potential Energy Well & Small Oscillations

Question:
A particle of mass $m=0.5\text{ kg}$ is moving in a 1D conservative force field whose potential energy function is given by:

$$U(x)=\frac{a}{x^2}-\frac{b}{x}$$

where $a=2\text{ J}\cdot {\text{m}}^2$ and $b=4\text{ J}\cdot \text{m}$.

1. Find the equilibrium position $x_0$.
2. Determine the binding energy $U_{min}$ of the particle.
3. Find the angular frequency $\omega$ for small oscillations about the stable equilibrium.

Step-by-Step Solution:

• 1. Equilibrium Position:

  $$F(x)=-\frac{dU}{dx}=-(-\frac{2a}{x^3}+\frac{b}{x^2})=\frac{2a-bx}{x^3}=0$$$$x_0=\frac{2a}{b}=\frac{2(2)}{4}=\boxed{{\displaystyle 1.0\text{ m}}}$$

• 2. Minimum Potential Energy:

  $$U(x_0)=\frac{a}{x_0^2}-\frac{b}{x_0}=\frac{2}{(1{)}^2}-\frac{4}{1}=2-4=\boxed{{\displaystyle -2.0\text{ J}}}$$

• 3. Oscillation Frequency:

  $$\frac{d^{2}U}{d{x}^2}=\frac{6a}{x^4}-\frac{2b}{x^3}$$

  At $x_0=1.0\text{ m}$:

  $$k_{\text{eff}}={\frac{d^{2}U}{d{x}^2}|}_{x=1}=\frac{6(2)}{1^4}-\frac{2(4)}{1^3}=12-8=4\text{ N/m}$$$$\omega =\sqrt{\frac{k_{\text{eff}}}{m}}=\sqrt{\frac{4}{0.5}}=\sqrt{8}=\boxed{{\displaystyle 2\sqrt{2}\text{ rad/s}\approx 2.83\text{ rad/s}}}$$

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PYQ 2: JEE Advanced 2018 — Variable Mass & Work Done by Chain

Question:
A uniform flexible chain of length $L$ and total mass $M$ is held vertically with its lowest link touching a table. The chain is released from rest. Find the total normal force exerted by the chain on the table as a function of the fallen distance $y$.

Step-by-Step Solution:

The total normal force $N(t)$ consists of two parts:

1. Weight of the chain already at rest on the table:$$W(y)=(\lambda y)g=\left(\frac{M}{L}y\right)g$$
2. Impulsive force from the continuous impact of the chain arriving at speed $v=\sqrt{2gy}$:$$F_{\text{impulse}}=v\frac{dm}{dt}=v(\lambda v)=\lambda v^2=\left(\frac{M}{L}\right)(2gy)=2\left(\frac{M}{L}y\right)g$$
3. Total Force on Table:$$N(y)=W(y)+F_{\text{impulse}}=\frac{Mgy}{L}+\frac{2Mgy}{L}=\boxed{{\displaystyle 3\left(\frac{M}{L}y\right)g=3W_{\text{resting}}}}$$

(The force on the table is exactly 3 times the resting weight of the fallen portion!)

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5. High-Yield Formula Sheet

Concept	Formula	Notes
Potential Energy Gradient	$\overrightarrow{F}=-\mathrm{\nabla }U$	Conservative fields only
Oscillation about Minimum	$\omega =\sqrt{\frac{U^{″}(x_0)}{m}}$	Stable equilibrium point
Spring Potential Energy	$U=\frac{1}{2}k{x}^2$	$W_{\text{spring}}=-\frac{1}{2}k(x_f^2-x_i^2)$
Looping the Loop (String)	$v_{min}=\sqrt{5gR}$	Tension at top $T\ge 0$
Looping the Loop (Rod)	$v_{min}=\sqrt{4gR}=2\sqrt{gR}$	Velocity at top $v\ge 0$