Why is potential energy only defined for conservative forces?

Potential energy $U$ represents stored configuration energy. A force is conservative if the work done by it around any closed loop is zero ($\oint \vec{F} \cdot d\vec{r} = 0$), making work strictly path-independent ($W = -\Delta U$). Non-conservative forces like friction dissipate mechanical energy as heat along path-dependent trajectories, so unique single-valued potential functions cannot exist for them.

What is the minimum speed needed at the bottom to complete a vertical circular loop?

For a mass tied to a light string of length $L$, the string must not go slack at the highest point, requiring minimum top speed $v_{\text{top}} = \sqrt{gL}$. Applying conservation of mechanical energy between top and bottom gives $E_{\text{bottom}} = E_{\text{top}} \implies \frac{1}{2}m v_{\text{bottom}}^2 = \frac{1}{2}m v_{\text{top}}^2 + mg(2L)$. Substituting $v_{\text{top}} = \sqrt{gL}$ yields $v_{\text{bottom}} = \sqrt{5gL}$.

Does the Work-Energy Theorem remain valid in non-inertial frames?

Yes, provided you include the work done by pseudo forces. In an accelerating frame of acceleration $\vec{a}_0$, pseudo force $\vec{F}_{\text{pseudo}} = -m\vec{a}_0$ acts on mass $m$. The generalized equation becomes $W_{\text{real}} + W_{\text{pseudo}} = \Delta K_{\text{rel}}$, where $\Delta K_{\text{rel}}$ is the kinetic energy change measured relative to the accelerating observer.

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Key Formulas & High-Yield Equations

Work Done by a Variable Force

$W = \int_{\vec{r}_1}^{\vec{r}_2} \vec{F} \cdot d\vec{r} = \int_{x_1}^{x_2} F_x dx + \int_{y_1}^{y_2} F_y dy + \int_{z_1}^{z_2} F_z dz$

Valid strictly in an inertial reference frame where force

\vec{F}

is expressed as a continuous function of position vector

\vec{r}

Work-Energy Theorem

$W_{\text{net}} = W_{\text{conservative}} + W_{\text{non-conservative}} + W_{\text{pseudo}} = \Delta K = K_f - K_i$

Applies to both rigid bodies and particle systems. In non-inertial frames, work done by pseudo forces

W_{\text{pseudo}}

must be included.

Conservative Force and Potential Energy Gradient

$\vec{F} = -\nabla U = -\left(\frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j} + \frac{\partial U}{\partial z}\hat{k}\right)$

Valid exclusively for conservative force fields. Potential energy

U

is undefined for dissipative non-conservative forces like friction.

Conservation of Mechanical Energy

$E_{\text{mech}} = K_i + U_i = K_f + U_f$

Valid strictly when the net work done by all non-conservative, dissipative, and external non-potential forces (

W_{\text{nc}} + W_{\text{ext}}

) is zero.

Instantaneous and Average Power

$P = \frac{dW}{dt} = \vec{F} \cdot \vec{v} = F v \cos\theta$

Valid for an instantaneous force

\vec{F}

acting on a body moving at instantaneous velocity

\vec{v}

relative to an inertial frame.

Coefficient of Restitution in 1D Collisions

$e = \frac{v_2 - v_1}{u_1 - u_2} = \frac{\text{Velocity of Separation}}{\text{Velocity of Approach}}$

Valid for one-dimensional impacts along the line of impact. Constraints:

e = 1

(perfectly elastic),

0 < e < 1

(inelastic),

e = 0

(perfectly inelastic).

Solved High-Yield Sample Questions

Practice Mode

Question #1NEET 2024

A particle of mass $m = 0.5\text{ kg}$ moves along the $x$ -axis under the action of a force $F = (3x^2 + 2x)\text{ N}$ . If the particle starts from rest at $x = 0$ , its kinetic energy at $x = 2\text{ m}$ is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Apply the Work-Energy Theorem, which states that net work done equals the change in kinetic energy: $W_{\text{net}} = \Delta K = K_f - K_i$ .

Step 2

Integrate the variable force from $x = 0\text{ m}$ to $x = 2\text{ m}$ to get work done: $W = \int_{0}^{2} (3x^2 + 2x) dx = \left[ x^3 + x^2 \right]_0^2 = (2^3 + 2^2) - 0 = 8 + 4 = 12\text{ J}$ .

Step 3

Since the particle starts from rest, $K_i = 0$ . Therefore, $K_f = W = 12\text{ J}$ .

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Question #2JEE Main 2023

A potential energy function for a conservative two-dimensional force field is given by $U(x, y) = 3x^2 y - y^3$ , where $U$ is in Joules and $x, y$ are in meters. The magnitude of the force acting on a particle at position $(1\text{ m}, 2\text{ m})$ is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Calculate force components using the gradient relation $\vec{F} = -\frac{\partial U}{\partial x}\hat{i} - \frac{\partial U}{\partial y}\hat{j}$ .

Step 2

Partial derivative with respect to $x$ gives $F_x = -\frac{\partial}{\partial x}(3x^2 y - y^3) = -6xy$ . At $(1, 2)$ , $F_x = -6(1)(2) = -12\text{ N}$ . Partial derivative with respect to $y$ gives $F_y = -\frac{\partial}{\partial y}(3x^2 y - y^3) = -(3x^2 - 3y^2) = 3y^2 - 3x^2$ . At $(1, 2)$ , $F_y = 3(2^2) - 3(1^2) = 12 - 3 = 9\text{ N}$ .

Step 3

Find net force magnitude: $|\vec{F}| = \sqrt{F_x^2 + F_y^2} = \sqrt{(-12)^2 + (9)^2} = \sqrt{144 + 81} = \sqrt{225} = 15\text{ N}$ .

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Mastering Work, Energy and Power: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Work Done by a Variable Force

Work-Energy Theorem

Conservative Force and Potential Energy Gradient

Conservation of Mechanical Energy

Instantaneous and Average Power

Coefficient of Restitution in 1D Collisions

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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