Table of Contents
How to Use This Page
Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This page completely covers Work, Energy and Power for CBSE Class 11 Physics.
Key Concepts
Work, Energy & Power
The currency of the physical universe
1
Work Done by a Constant Force
Formula
Work is the dot product of Force and Displacement. It is a scalar quantity. Work is maximum when $\theta = 0^\circ$, zero when $\theta = 90^\circ$, and negative when $\theta > 90^\circ$.
$$W = \vec{F} \cdot \vec{s} = Fs \cos \theta$$
2
Work Done by a Variable Force
Calculus
When the force changes with position (like a spring force), work is calculated by integrating the force over the displacement. Graphically, it is the Area under the Force-Displacement ($F$-$x$) curve.
$$W = \int_{x_i}^{x_f} \vec{F} \cdot d\vec{x}$$
3
Kinetic Energy & Momentum Relation
Formula
Kinetic energy ($K$) is the energy possessed by a body due to its motion. It is directly related to the square of its linear momentum ($p$).
$$K = \frac{1}{2}mv^2 = \frac{p^2}{2m} \quad \implies \quad p = \sqrt{2mK}$$
4
Work-Energy Theorem
Formula
The net work done by ALL forces (conservative, non-conservative, and external) acting on a body is exactly equal to the change in its kinetic energy.
$$W_{\text{net}} = \Delta K = K_f – K_i$$
5
Potential Energy (Gravitational & Elastic)
Formula
Energy stored due to position or configuration. Change in potential energy ($\Delta U$) is defined as the negative of work done by conservative forces.
$$U_{\text{grav}} = mgh$$
$$U_{\text{spring}} = \frac{1}{2}kx^2 \quad \text{($k$ is spring constant)}$$
6
Conservation of Mechanical Energy
Formula
If only conservative forces (like gravity or spring force) are doing work on a system, the total mechanical energy ($E = K + U$) remains constant.
$$K_i + U_i = K_f + U_f \quad \implies \quad \Delta K + \Delta U = 0$$
7
Power
Formula
The rate at which work is done or energy is transferred. Instantaneous power can be expressed as the dot product of Force and Velocity. SI Unit: Watt (W).
$$P_{\text{avg}} = \frac{W}{\Delta t} \quad | \quad P_{\text{inst}} = \frac{dW}{dt} = \vec{F} \cdot \vec{v}$$
8
Coefficient of Restitution ($e$)
Formula
The ratio of the relative velocity of separation after collision to the relative velocity of approach before collision. It measures the “bounciness” of the collision.
$$e = \frac{v_2 – v_1}{u_1 – u_2} \quad (0 \le e \le 1)$$
9
Velocities in 1D Elastic Collision
Formula
When $e=1$, both momentum and kinetic energy are conserved. The final velocities of masses $m_1$ and $m_2$ can be directly calculated using these standard derivations:
$$v_1 = \left(\frac{m_1 – m_2}{m_1 + m_2}\right)u_1 + \left(\frac{2m_2}{m_1 + m_2}\right)u_2$$
$$v_2 = \left(\frac{2m_1}{m_1 + m_2}\right)u_1 + \left(\frac{m_2 – m_1}{m_1 + m_2}\right)u_2$$
Concept Deep Dive
01
The Work-Energy Theorem
The ultimate physics cheat codeIf a problem asks you for final velocity, and you have distance and forces (like friction, gravity, applied pull), your instinct might be to use Newton’s Laws: find net force, find acceleration, use $v^2 = u^2 + 2as$.
Skip all that. The Work-Energy Theorem ($W_{\text{net}} = \Delta K$) bypasses time and acceleration entirely. Just calculate the work done by gravity, add the work done by friction (which will be negative), and set the total equal to $\frac{1}{2}mv_f^2 – \frac{1}{2}mv_i^2$. It is faster, less prone to algebra errors, and works even if the forces are variable!
Skip all that. The Work-Energy Theorem ($W_{\text{net}} = \Delta K$) bypasses time and acceleration entirely. Just calculate the work done by gravity, add the work done by friction (which will be negative), and set the total equal to $\frac{1}{2}mv_f^2 – \frac{1}{2}mv_i^2$. It is faster, less prone to algebra errors, and works even if the forces are variable!
02
Coefficient of Restitution ($e$)
How much energy survives the crash?The value of $e$ tells you exactly what kind of collision happened:
$e = 1$ (Perfectly Elastic): Objects bounce off perfectly. Kinetic energy is 100% conserved. (e.g., Ideal gas molecules, billiard balls).
$0 < e < 1$ (Inelastic): Objects bounce, but some kinetic energy is permanently lost to heat, sound, or deformation. (e.g., A tennis ball hitting the floor).
$e = 0$ (Perfectly Inelastic): Objects stick together after crashing and move with a common final velocity. Maximum possible kinetic energy is lost. (e.g., A bullet embedding into a wooden block).
$e = 1$ (Perfectly Elastic): Objects bounce off perfectly. Kinetic energy is 100% conserved. (e.g., Ideal gas molecules, billiard balls).
$0 < e < 1$ (Inelastic): Objects bounce, but some kinetic energy is permanently lost to heat, sound, or deformation. (e.g., A tennis ball hitting the floor).
$e = 0$ (Perfectly Inelastic): Objects stick together after crashing and move with a common final velocity. Maximum possible kinetic energy is lost. (e.g., A bullet embedding into a wooden block).
Compare & Contrast
✗ Conservative Forces
- Work done depends ONLY on initial and final positions.
- Work done in a closed loop is exactly zero.
- Mechanical energy ($K + U$) is conserved.
- Examples: Gravity, Electrostatic Force, Spring Force.
✓ Non-Conservative Forces
- Work done depends heavily on the path taken.
- Work done in a closed loop is not zero (it dissipates energy).
- Mechanical energy is lost (usually as heat or sound).
- Examples: Friction, Air Resistance, Viscous Drag.
Remember
Potential energy ($U$) can ONLY be defined for Conservative forces. You can have “Gravitational Potential Energy”, but there is no such thing as “Frictional Potential Energy”!
Common Mistakes to Avoid
Mistake 1
Assuming Work is always $F \times d$: If you are given a graph of Force vs. Position, or if the force is an equation like $F = 3x^2$, you cannot just multiply. You MUST find the area under the graph or integrate the function ($W = \int F dx$).
Mistake 2
Forgetting Work can be Negative: Work is a scalar, but it can absolutely be negative if the angle $\theta$ is greater than $90^\circ$. Friction always does negative work because it acts at $180^\circ$ to the displacement ($\cos 180^\circ = -1$). Negative work means energy is being *removed* from the object.
Mistake 3
Applying Conservation of Mechanical Energy blindly: You cannot use $K_i + U_i = K_f + U_f$ if there is friction on the surface! If friction is present, you must use the modified equation: $K_i + U_i + W_{\text{friction}} = K_f + U_f$.
Exam Tips
Tip 1
Percentage Change Shortcut: Since $K = p^2/2m$, if momentum $p$ increases by a very small amount $x\%$ (where $x < 5\%$), Kinetic Energy $K$ will increase by approximately $2x\%$. (Example: If momentum increases by $2\%$, KE increases by $\approx 4\%$). This relies on binomial expansion.
Tip 2
Special Case in Elastic Collisions: If two objects of equal mass collide completely elastically in 1D, they simply swap velocities! (If $m_1 = m_2$, then $v_1 = u_2$ and $v_2 = u_1$). This saves doing the massive derivation in MCQ exams.
Expected Exam Questions
SQ
Board Pattern Questions
Class 11 · Work, Energy & Power · CBSE Exam1
A coolie carrying a heavy load of $50 \text{ kg}$ on his head walks $100 \text{ m}$ on a perfectly horizontal platform. What is the work done by him against gravity? [1 mark]
Answer
Zero
📝
Explanation
The force of gravity acts vertically downwards ($mg$), and the displacement is strictly horizontal. The angle between the force and displacement is $\theta = 90^\circ$.
Work done $W = Fs \cos(90^\circ) = F \times s \times 0 = 0 \text{ Joules}$.
2
A force $F = (3x^2 + 2x – 5) \text{ N}$ acts on a particle. Calculate the work done in displacing the particle from $x = 1 \text{ m}$ to $x = 3 \text{ m}$. [2 marks]
Answer
$24 \text{ Joules}$
📝
Explanation
Because the force varies with position $x$, we must integrate:
$W = \int_{1}^{3} (3x^2 + 2x – 5) dx$
$W = \left[ \frac{3x^3}{3} + \frac{2x^2}{2} – 5x \right]_{1}^{3} = [x^3 + x^2 – 5x]_{1}^{3}$
Upper limit: $(3^3 + 3^2 – 5(3)) = 27 + 9 – 15 = 21$
Lower limit: $(1^3 + 1^2 – 5(1)) = 1 + 1 – 5 = -3$
$W = 21 – (-3) = 21 + 3 = 24 \text{ J}$.
3
A bullet of mass $0.012 \text{ kg}$ and horizontal speed $70 \text{ m/s}$ strikes a wooden block of mass $0.4 \text{ kg}$ resting on a frictionless table, and embeds itself in it. Find the kinetic energy lost in the collision. Where does this energy go? [3 marks]
Answer
Lost KE $\approx 28.5 \text{ J}$. Converted to heat/sound.
📝
Explanation
This is a perfectly inelastic collision ($e=0$).
1. Find final common velocity ($v$) using Conservation of Momentum:
$m_1 u_1 + m_2 u_2 = (m_1 + m_2)v \implies (0.012)(70) + 0 = (0.012 + 0.4)v \implies 0.84 = 0.412 v \implies v \approx 2.04 \text{ m/s}$.
2. Calculate Initial KE: $K_i = \frac{1}{2}m_1 u_1^2 = \frac{1}{2}(0.012)(70)^2 = 0.006 \times 4900 = 29.4 \text{ J}$.
3. Calculate Final KE: $K_f = \frac{1}{2}(m_1+m_2)v^2 = \frac{1}{2}(0.412)(2.04)^2 \approx 0.86 \text{ J}$.
4. Energy Lost: $\Delta K = K_i – K_f = 29.4 – 0.86 = 28.54 \text{ J}$.
This massive loss of kinetic energy is dissipated as heat, sound, and the mechanical work done to tear through the wood fibers.
Concept Map
Work, Energy & Power connects to →
Mechanics
Newton’s Laws
Conservation Principles
Thermodynamics (Heat loss)
System of Particles
Spring Dynamics
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