Work Energy and Power revision notes

Key Concepts & Definitions

Work, energy, and power are fundamental concepts in physics with very precise definitions, unlike their everyday usage.

Work (WWW):

Work is said to be done by a force on a body over a certain displacement if the force has a component in the direction of the displacement,.

Kinetic Energy (KKK):

The capacity of an object to do work by virtue of its motion. It is defined as half the mass times the square of the speed,.

Potential Energy (VVV):

The stored energy by virtue of the position or configuration of a body. It is only defined for conservative forces.

Conservative Force:

A force is conservative if the work it does on an object depends only on the initial and final positions (path-independent). Alternatively, the work done by a conservative force in any closed path is exactly zero,. Examples include gravitational and spring forces,.

Non-Conservative Force:

Forces for which the work done depends on the path taken (e.g., friction, viscous drag),. No potential energy can be defined for these forces.

Mechanical Energy (EEE):

The sum of the kinetic and potential energies of a system (E=K+VE = K + VE=K+V).

Power (PPP):

The time rate at which work is done or energy is transferred.

Collisions:

An interaction where mutual impulsive forces act over a short collision time, causing a change in momenta.

Elastic Collision:

A collision where deformation is entirely relieved, meaning both total linear momentum and total kinetic energy are conserved.

Inelastic Collision:

A collision where some initial kinetic energy is transformed into heat or sound; only total linear momentum is conserved.

Completely Inelastic Collision:

A collision where the colliding bodies do not relieve deformation and move together with a common velocity after the impact.

Mathematical Prerequisite: The Scalar (Dot) Product

Physical quantities like force and displacement are vectors, and their combination to calculate work requires the scalar product,.

• The scalar (dot) product of vectors $\mathbf{A}$ and $\mathbf{B}$ is a scalar quantity defined as $\mathbf{A} \cdot \mathbf{B} = AB \cos\theta$, where $\theta$ is the angle between the two vectors.
• Geometrical meaning: It is the product of the magnitude of $\mathbf{A}$ and the component of $\mathbf{B}$ along $\mathbf{A}$ ($B \cos\theta$), or vice versa.
• Properties:
  • Commutative Law: $\mathbf{A} \cdot \mathbf{B} = \mathbf{B} \cdot \mathbf{A}$.
  • Distributive Law: $\mathbf{A} \cdot (\mathbf{B} + \mathbf{C}) = \mathbf{A} \cdot \mathbf{B} + \mathbf{A} \cdot \mathbf{C}$.
  • Scalar multiplication: $\mathbf{A} \cdot (\lambda \mathbf{B}) = \lambda (\mathbf{A} \cdot \mathbf{B})$.
• Orthogonal Vectors: $\mathbf{A} \cdot \mathbf{B} = 0$ if vectors $\mathbf{A}$ and $\mathbf{B}$ are perpendicular ($\cos 90^\circ = 0$).
• Unit Vectors: $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$ and $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$,.
• Component Form: For $\mathbf{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$ and $\mathbf{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}$, the dot product is $\mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z$.
• Self Dot Product: $\mathbf{A} \cdot \mathbf{A} = A^2 = A_x^2 + A_y^2 + A_z^2$.
• JEE TIPTo find the angle between force and displacement, always use $\cos\theta = \frac{\mathbf{F} \cdot \mathbf{d}}{|\mathbf{F}| |\mathbf{d}|}$,.

Work

Work is defined as the product of the component of the force in the direction of the displacement and the magnitude of this displacement,.

• Constant Force: $W = (\mathbf{F} \cos\theta)d = \mathbf{F} \cdot \mathbf{d}$.
• Variable Force: If force varies with position $F(x)$, work is the integral of force over displacement: $W = \int_{x_i}^{x_f} F(x) dx$.
• Zero Work: Work is strictly zero if: (i) displacement is zero (e.g., pushing a rigid wall, holding a weight steadily), (ii) force is zero, or (iii) force and displacement are mutually perpendicular ($\theta = 90^\circ$),.
• Positive/Negative Work: Work is positive for acute angles ($0^\circ \le \theta < 90^\circ$) and negative for obtuse angles ($90^\circ < \theta \le 180^\circ$). Kinetic friction typically does negative work as it opposes relative motion ($\theta = 180^\circ$),.

Kinetic Energy & The Work-Energy Theorem

The kinetic energy $K$ is a scalar quantity.

• $K = \frac{1}{2} m \mathbf{v} \cdot \mathbf{v} = \frac{1}{2} m v^2$.
• Work-Energy (WE) Theorem: The change in kinetic energy of a particle is equal to the work done on it by the net force.
• Equation: $K_f - K_i = W_{net}$,.
• The WE theorem holds for constant forces, and is proven for variable forces by integrating Newton's second law: $\int_{x_i}^{x_f} F dx = \int_{v_i}^{v_f} mv dv = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$,,.
• JEE TIPThe WE theorem is an integral form of Newton's second law; temporal (time) information is integrated out, meaning you can find speeds at certain positions, but NOT the time taken to travel.

Potential Energy & Conservation of Mechanical Energy

Potential energy $V(x)$ is defined strictly for conservative forces such that $\Delta V = - \int_{x_i}^{x_f} F(x) dx$, implying $F(x) = -\frac{dV}{dx}$.

• Gravitational Potential Energy: $V(h) = mgh$, where the zero of potential energy is conventionally at the earth's surface ($h=0$),.
• Conservation of Mechanical Energy: For a conservative force $\Delta K = -\Delta V$, which yields $\Delta (K + V) = 0$. The sum $E = K + V$ remains strictly constant throughout the motion.
• When non-conservative forces (like friction) are present, the WE theorem becomes $W_c + W_{nc} = \Delta K$, leading to $W_{nc} = \Delta E = E_f - E_i$.

Potential Energy of a Spring

The spring force is an ideal example of a variable, conservative force,.

• Hooke’s Law: $F_s = -kx$, where $x$ is displacement from equilibrium and $k$ is the spring constant ($N/m$),.
• Work done by spring force: $W_s = \int_{x_i}^{x_f} (-kx) dx = \frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2$.
• Elastic Potential Energy: $V(x) = \frac{1}{2}kx^2$, defined such that $V(0) = 0$ at the unstretched equilibrium position.
• Total mechanical energy of a horizontal spring-mass system: $E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \text{constant}$. Maximum speed $v_m = x_m \sqrt{\frac{k}{m}}$ occurs at equilibrium ($x=0$).

Power

Power is the rate at which work is done.

• Average Power: $P_{av} = \frac{W}{t}$.
• Instantaneous Power: $P = \frac{dW}{dt} = \mathbf{F} \cdot \frac{d\mathbf{r}}{dt} = \mathbf{F} \cdot \mathbf{v}$.
• Fluid Power & Mass Flow Rate (Windmills & Pumps): When dealing with continuous mass flow (like air through a windmill or water from a pump), the mass passing through an area $A$ in time $t$ with velocity $v$ is $m = \rho A v t$, where $\rho$ is the fluid density. The kinetic energy available is $K = \frac{1}{2}mv^2 = \frac{1}{2}(\rho A v t)v^2$. Therefore, the power available from the wind/fluid is $P = \frac{K}{t} = \frac{1}{2}\rho A v^3$.
• Kilowatt-hour (kWh): It is crucial to remember that the kilowatt-hour is a unit of energy, not power,. $1 \text{ kWh} = 3.6 \times 10^6 \text{ J}$.

Collisions

In collisions, mutual impulsive forces dictate motion. According to Newton's Third Law, total impulse is zero, conserving the system's total linear momentum.

• One-Dimensional (Head-On) Elastic Collision:
  • Consider mass $m_1$ (velocity $v_{1i}$) colliding with mass $m_2$ at rest ($v_{2i} = 0$),.
  • Momentum conservation: $m_1 v_{1i} = m_1 v_{1f} + m_2 v_{2f}$.
  • Kinetic energy conservation: $\frac{1}{2}m_1 v_{1i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2$.
  • Final velocities: $v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v_{1i}$ and $v_{2f} = \frac{2m_1}{m_1 + m_2} v_{1i}$,.
  • Special Case 1 ($m_1 = m_2$): $v_{1f} = 0$ and $v_{2f} = v_{1i}$ (Velocities are exchanged).
  • Special Case 2 ($m_2 \gg m_1$): $v_{1f} \approx -v_{1i}$ and $v_{2f} \approx 0$ (Lighter mass bounces back with same speed).
• One-Dimensional Completely Inelastic Collision:
  • Masses stick together: $v_f = \frac{m_1 v_{1i}}{m_1 + m_2}$.
  • Loss in kinetic energy: $\Delta K = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} v_{1i}^2$.
• Two-Dimensional (Oblique) Elastic Collision:
  • Conservation equations must be written for x and y axes: $m_1 v_{1i} = m_1 v_{1f} \cos\theta_1 + m_2 v_{2f} \cos\theta_2$ and $0 = m_1 v_{1f} \sin\theta_1 - m_2 v_{2f} \sin\theta_2$,.
• Scattering (Action at a Distance): Collisions do not strictly require physical contact. Phenomena like a comet approaching the sun or an alpha particle deflected by a nucleus involve forces acting at a distance. These events are termed "scattering", and conservation laws apply identically based on the initial and final states far from the interaction region.

Conditions & Limitations

• Gravitational potential energy $V = mgh$ is strictly valid ONLY near the earth's surface where variation of $g$ is negligible ($h \ll R_E$),.
• Mechanical Energy ($K+V$) conservation is ONLY valid when strictly conservative forces are doing work. If friction or air drag is present, you must use $W_{nc} = E_f - E_i$.
• The WE Theorem relies on inertial reference frames. It can be extended to non-inertial frames ONLY if pseudo-forces are explicitly added to the calculation of net work.

Important Graphs & Diagrams

• Varying Force Graph: A plot of force $F(x)$ versus displacement $x$. The area under the curve enclosed between $x_i$ and $x_f$ represents the exact work done by the variable force. Areas below the x-axis (negative force) contribute negative work,.
• Energy vs Displacement for Spring: Parabolic plots of $V = \frac{1}{2}kx^2$ (upward opening parabola) and $K = E - \frac{1}{2}kx^2$ (downward opening parabola),. They intersect at $x = \pm \frac{x_m}{\sqrt{2}}$. The total energy $E$ is a horizontal straight line, showing $K + V$ is constant.
• Potential Energy Curves & Turning Points: For a given potential energy curve $V(x)$, the total energy $E$ is represented by a horizontal line since $E$ is constant.
  • Forbidden Regions: A particle can never be found in regions where $V(x) > E$ because kinetic energy ($K = E - V$) cannot be negative.
  • Turning Points: The points where the $E$ line intersects the $V(x)$ curve are "turning points" because $K=0$ and the particle must reverse its direction.

Standard Derivations & Step-by-Step Problem Solving

1. Work-Energy Theorem for Variable Force

• Step 1: Start with the time rate of change of kinetic energy: $\frac{dK}{dt} = \frac{d}{dt}(\frac{1}{2}mv^2) = mv \frac{dv}{dt}$.
• Step 2: From Newton's Second Law, $m\frac{dv}{dt} = F$, and velocity $v = \frac{dx}{dt}$.
• Step 3: Substitute these to get $\frac{dK}{dt} = F \frac{dx}{dt}$, which simplifies to $dK = F dx$.
• Step 4: Integrate from initial position $x_i$ to final position $x_f$: $\int_{K_i}^{K_f} dK = \int_{x_i}^{x_f} F dx \implies K_f - K_i = W$.

2. Vertical Circular Motion (String Bob)

• Scenario: A bob of mass $m$ suspended by string $L$, given horizontal velocity $v_0$ at the lowest point A. Goal: Reach highest point C with string going slack exactly at C.
• Step 1: At highest point C, tension $T_c \to 0$. Net inward force is gravity. $mg = \frac{mv_c^2}{L} \implies v_c = \sqrt{gL}$.
• Step 2: Energy at C: $E_c = \frac{1}{2}m(\sqrt{gL})^2 + mg(2L) = \frac{5}{2}mgL$.
• Step 3: Conserve energy between lowest point A ($V_A=0$) and C: $\frac{1}{2}mv_0^2 = \frac{5}{2}mgL \implies v_0 = \sqrt{5gL}$.
• Step 4: Speed at horizontal point B ($V_B = mgL$): $\frac{1}{2}mv_B^2 + mgL = \frac{5}{2}mgL \implies v_B = \sqrt{3gL}$.
• JEE TIPIf the string breaks or goes completely slack at point C, the bob immediately executes a parabolic projectile motion since it has a purely horizontal velocity.

3. Moderation of Neutrons (Collision Energy Transfer)

• Scenario: Fast neutron (mass $m_1$) elastically collides with stationary target nucleus (mass $m_2$),.
• Step 1: Initial $K_i = \frac{1}{2}m_1 v_{1i}^2$.
• Step 2: Final speed $v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v_{1i}$.
• Step 3: Fractional kinetic energy retained by neutron $f_1 = \frac{K_{1f}}{K_{1i}} = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)^2$.
• Step 4: Fractional kinetic energy transferred to target $f_2 = 1 - f_1 = \frac{4m_1 m_2}{(m_1 + m_2)^2}$. Maximum energy is transferred when $m_1 \approx m_2$ (e.g., Deuterium or Carbon moderators in nuclear reactors),.

Previous Year JEE Topics

Based on the foundational concepts in this chapter, the most frequently tested mechanics topics in JEE include:

• Integration of Variable Forces: Calculating work done using $\int \mathbf{F} \cdot d\mathbf{r}$ specifically when force is a function of $x$, $y$, or time,.
• Work-Energy Theorem with Friction: Problems involving moving masses on rough patches where $\Delta K = W_{gravity} + W_{spring} + W_{friction}$.
• Vertical Circular Motion: Checking conditions for looping the loop ($v \ge \sqrt{5gL}$), pendulum oscillations ($v \le \sqrt{2gL}$), and string slackening causing projectile motion,.
• Coefficient of Restitution & Oblique Collisions: Applying momentum conservation along the axis of collision and preserving velocity tangential to the collision surface,. Specially tested is the $90^\circ$ separation of equal masses after an elastic glancing collision,.
• Power Kinematics: Finding the proportional relationships $v \propto \sqrt{t}$ and $x \propto t^{3/2}$ for machines delivering constant power.

JEE Traps

• [JEE TIP] Trap 1 - The Force-Displacement Zero Work Blindspot:

  • Misconception: If the net mechanical work evaluated over a path is zero ($W = \mathbf{F} \cdot \mathbf{d} = 0$), it guarantees that either the applied force vector or the displacement vector must have collapsed to zero.
  • Correct Understanding: Work is a scalar dot product ($W = Fd\cos\theta$). It drops to exactly zero whenever the net force vector operates perfectly perpendicular ($\theta = 90^\circ$) to the instantaneous displacement vector. Classic examples include the centripetal gravitational force keeping a satellite in a stable circular orbit, or the normal reaction force acting on a block sliding horizontally across a flat floor.

• [JEE TIP] Trap 2 - The 2D Elastic Glancing Angle Monopoly:

  • Misconception: Any two masses undergoing a two-dimensional elastic oblique collision will automatically fly away along trajectories oriented at a perfect $90^\circ$ angle relative to each other.
  • Correct Understanding: The post-collision orthogonal split ($\theta_1 + \theta_2 = 90^\circ$) is a highly restricted boundary case. It holds true if and only if both colliding masses are perfectly identical ($m_1 = m_2$) AND the target mass was strictly at rest ($\mathbf{v}_2 = \mathbf{0}$) prior to impact. If the masses differ or the target is moving, the scattering angle will deviate from $90^\circ$.

• [JEE TIP] Trap 3 - Constant Power Kinematic Scaling:

  • Misconception: A vehicle accelerating from rest under a perfectly constant engine power output will experience a uniform, constant acceleration, causing its velocity to scale linearly with time ($v \propto t$).
  • Correct Understanding: Because power is the product of force and velocity ($P = Fv = mv\frac{dv}{dt}$), a constant power constraint dictates that force must decrease as velocity climbs. Separating variables and integrating yields a non-linear velocity scaling profile: $v \propto \sqrt{t}$. Further integration shows that displacement scales as $x \propto t^{3/2}$, meaning the vehicle's acceleration actually decreases over time ($\mathbf{a} \propto t^{-1/2}$).

• [JEE TIP] Trap 4 - Continuous Collision Energy Conservation:

  • Misconception: In a perfectly elastic collision, the total kinetic energy of the system is perfectly conserved at every single microsecond of the interaction, including the brief duration of physical contact.
  • Correct Understanding: Total kinetic energy is conserved strictly when comparing the isolated initial and final steady states ($K_i = K_f$). During the intermediate impact window, kinetic energy is strictly not conserved. At the point of maximum deformation, a significant fraction of the initial kinetic energy is temporarily converted and stored as elastic potential energy within the deformed structures before being released during the restitution phase.

• [JEE TIP] Trap 5 - Internal Action-Reaction Work Cancellation:

  • Misconception: Because Newton's Third Law dictates that internal action and reaction forces are always equal and opposite, the net work performed by all internal forces within a closed system must sum to zero.
  • Correct Understanding: While the forces are equal and opposite ($\mathbf{F}_{12} = -\mathbf{F}_{21}$), mechanical work depends on the individual displacement vectors of each point of application ($W = \mathbf{F} \cdot \mathbf{d}$). If the two interacting bodies undergo different physical displacements, the net internal work is non-zero ($W_{12} + W_{21} \neq \mathbf{0}$). Consequently, internal forces routinely alter a system's internal kinetic energy, as seen in explosions, spring releases, or inelastic deformations.

• [JEE TIP] Trap 6 - The Leaking Sand Cart Momentum Illusion:

  • Misconception: If a cargo cart coasting at a constant speed along a frictionless track begins leaking sand through a hole in its floor, the cart will speed up or slow down to balance its changing mass.
  • Correct Understanding: The cart's horizontal velocity remains completely unchanged ($\Delta v = 0$). As the sand drains, it leaves the cart with an identical horizontal velocity component, meaning it exerts zero horizontal reaction thrust back on the vehicle. While the total system momentum ($p = mv$) drops linearly because the mass of the cart is decreasing, the scalar speed stays perfectly locked.

• [JEE TIP] Trap 7 - The Kilowatt-Hour Dimensional Identity:

  • Misconception: The kilowatt-hour ($\text{kWh}$) is an exceptionally large industrial unit utilized to quantify the raw power output or mechanical capacity of heavy machinery.
  • Correct Understanding: The kilowatt-hour is strictly a unit of Energy (Work), not power. It represents the product of a power unit (kilowatts) and a time unit (hours): $1\text{ kWh} = (1000\text{ J/s}) \times (3600\text{ s}) = \mathbf{3.6 \times 10^6\text{ Joules}}$. Mixing this up inside energy efficiency calculations leads to massive dimensional errors.

• [JEE TIP] Trap 8 - Closed Loop Conservative Monopoly:

  • Misconception: The net mechanical work performed by any physical force vector evaluated over a complete, closed geometric loop is universally equal to zero.
  • Correct Understanding: This vanishing work constraint applies exclusively to conservative forces (such as gravity, electrostatic forces, or ideal spring mechanics). For non-conservative forces like friction or air drag, the force vector continuously flips its orientation to directly oppose the instantaneous direction of motion, ensuring that the work integral accumulates a net negative scalar value proportional to the total path length ($\oint \mathbf{F}_{\text{fric}} \cdot d\mathbf{r} < 0$).

• [JEE TIP] Trap 9 - Work-Energy Theorem Boundary Limits:

  • Misconception: The foundational Work-Energy Theorem equation ($W = \Delta K$) is a simplified model that holds true if and only if all interacting forces within the system are strictly conservative.
  • Correct Understanding: The Work-Energy Theorem ($W_{\text{net}} = \Delta K$) is a universal law of mechanics with zero exceptions. The only constraint is that the leading variable $W_{\text{net}}$ must rigorously incorporate the individual scalar work contributions performed by every single active force acting on the object, including conservative forces, non-conservative forces (friction), and external driving agents.

• [JEE TIP] Trap 10 - Massive Impact Energy Transfer Fallacy:

  • Misconception: In a one-dimensional elastic collision, a massively heavy moving projectile will transfer the maximum possible fraction of its kinetic energy to a tiny, stationary target block.
  • Correct Understanding: A massive projectile retains almost all of its initial kinetic energy and plows forward virtually undeflected, transferring very little energy. Conversely, a tiny projectile will simply bounce back off a massive target with nearly identical speed, also leaving the target stationary. The fractional energy transfer hits its absolute maximum ($100\%$ transfer) if and only if the masses are perfectly equal ($m_1 = m_2$), causing the moving projectile to come to a dead stop and pass its entire energy profile to the target.