Work, Energy and Power

In physics, work is said to be done by a force when the point of application of the force undergoes a displacement having a non-zero component along the direction of the force. If a constant force $\vec{F}$ acts on a body and displaces it by $\vec{s}$, the work done $W$ is mathematically defined as the scalar (dot) product of the force vector and displacement vector:

where $F = |\vec{F}|$, $s = |\vec{s}|$, and $\theta$ is the angle between the force and displacement vectors ($0 \le \theta \le \pi$).

In Cartesian component form, if $\vec{F} = F_x\hat{i} + F_y\hat{j} + F_z\hat{k}$ and $\vec{s} = \Delta x\hat{i} + \Delta y\hat{j} + \Delta z\hat{k}$:

• Positive Work ($0 \le \theta < 90^\circ$, $\cos\theta > 0$): Force has a component along displacement. Example: Gravity acting on a falling body; an engine pulling a train forward.
• Negative Work ($90^\circ < \theta \le 180^\circ$, $\cos\theta < 0$): Force opposes displacement. Example: Kinetic friction acting on a sliding crate ($W_f = -f_k s$); gravitational force when lifting a body upward ($W_g = -mgh$).
• Zero Work ($W = 0$): Occurs under three distinct physical scenarios:
  1. Zero displacement ($s = 0$): Pushing against a rigid stationary wall ($W = 0$).
  2. Zero force ($F = 0$): A body moving with uniform velocity in deep space where net external force is zero.
  3. Perpendicular force and displacement ($\theta = 90^\circ$, $\cos 90^\circ = 0$): Centripetal force acting on a planet in circular orbit; normal reaction force on a body sliding across a horizontal table; a porter carrying a load on his head while walking horizontally ($W_g = mg \cdot s \cdot \cos 90^\circ = 0$).

In realistic scenarios, forces vary continuously with position (e.g., spring forces, gravitational forces, electrostatic forces). To compute the work done by a position-dependent force $\vec{F}(\vec{r})$ as the body moves from $\vec{r}_i$ to $\vec{r}_f$, we divide the continuous trajectory into infinitesimally small displacement elements $d\vec{r}$:

The Work-Energy Theorem states: The net work done by all external forces acting on a body equals the net change in its kinetic energy.

Rigorous Mathematical Proof:

By definition, kinetic energy is $K = \frac{1}{2}mv^2$. Differentiating both sides with respect to time $t$:

From Newton's Second Law, $F = ma$. Furthermore, velocity is $v = \frac{dx}{dt}$. Substituting these yields:

Integrating from the initial state ($x = x_i, v = v_i, K = K_i$) to the final state ($x = x_f, v = v_f, K = K_f$):

This scalar theorem is universally valid in all inertial reference frames and holds regardless of whether the forces are constant or variable, conservative or non-conservative.

Kinetic energy is the capacity of a material body to perform work by virtue of its state of motion. For a body of mass $m$ moving with speed $v$:

• Scalar Nature: Kinetic energy is strictly a scalar quantity and is always non-negative ($K \ge 0$) for non-zero mass.
• SI Unit: Joule ($\text{J} = \text{kg}\cdot\text{m}^2/\text{s}^2 = \text{N}\cdot\text{m}$). CGS unit: $\text{erg} = 10^{-7}\,\text{J}$.
• Relation with Linear Momentum: Since $p = mv$, we have: $$K = \frac{(mv)^2}{2m} = \frac{p^2}{2m} \implies p = \sqrt{2mK}$$ This fundamental relation is widely tested in board exams. For two bodies of masses $m_1$ and $m_2$ having identical momentum, $K \propto 1/m$ (the lighter body has greater kinetic energy). If they have identical kinetic energy, $p \propto \sqrt{m}$ (the heavier body possesses greater momentum).

Potential energy ($U$) is the energy possessed by a body or system by virtue of its position, configuration, or spatial arrangement in a force field. It is defined exclusively for conservative forces as the negative of the work done by the conservative force in moving the body from a standard reference level to its current position:

For uniform gravitational field near Earth's surface (taking $U = 0$ at ground level $h = 0$):

Consider an ideal, massless spring attached horizontally to a rigid wall and a block of mass $m$ on a frictionless table. According to Hooke's Law, for small displacements $x$ from the unextended equilibrium position ($x = 0$), the restoring spring force $F_s$ is linearly proportional and opposite to displacement:

where $k$ is the spring constant (stiffness factor) with SI units $\text{N/m}$ and dimensions $[M^1 T^{-2}]$.

The work done by an external agent $F_{\text{ext}} = +kx$ against the spring force to stretch the spring slowly from $x = 0$ to extension $x$ is stored as elastic potential energy:

Alternatively, using the conservative force definition $\Delta U = -W_s$:

• The potential energy $U_s = \frac{1}{2}kx^2$ is strictly positive whether the spring is stretched ($x > 0$) or compressed ($x < 0$).
• Work Done in Changing Extension from $x_1$ to $x_2$: $$W_{\text{spring}} = -\Delta U_s = \frac{1}{2}k x_1^2 - \frac{1}{2}k x_2^2$$ $$W_{\text{ext}} = \Delta U_s = \frac{1}{2}k (x_2^2 - x_1^2)$$

If non-conservative forces (like friction $f_k$) perform work $W_{\text{nc}}$, the generalized work-energy relation becomes:

Consider a body of mass $m$ released from rest from a height $h$ above the ground under gravity ($g$). We evaluate the total mechanical energy $E = K + U$ at three benchmark elevations:

1. At Position A (Initial point at height $h$):

   Speed $v_A = 0 \implies K_A = 0$. Potential energy $U_A = mgh$.

   $$E_A = K_A + U_A = 0 + mgh = mgh \quad \text{--- (1)}$$
2. At Position B (Intermediate point after falling distance $x$, at height $h - x$):

   Using third equation of kinematics $v_B^2 = u^2 + 2gx = 0 + 2gx = 2gx$.

   Kinetic energy: $K_B = \frac{1}{2}m v_B^2 = \frac{1}{2}m(2gx) = mgx$.

   Potential energy: $U_B = mg(h - x) = mgh - mgx$.

   $$E_B = K_B + U_B = mgx + (mgh - mgx) = mgh \quad \text{--- (2)}$$
3. At Position C (Just before striking the ground, height $= 0$):

   Falling total distance $h$: $v_C^2 = 0 + 2gh = 2gh$.

   Kinetic energy: $K_C = \frac{1}{2}m v_C^2 = \frac{1}{2}m(2gh) = mgh$.

   Potential energy: $U_C = mg(0) = 0$.

   $$E_C = K_C + U_C = mgh + 0 = mgh \quad \text{--- (3)}$$

Comparing (1), (2), and (3):

As the body falls, potential energy continuously transforms into kinetic energy, while the total mechanical energy remains strictly conserved at every instant.

Motion in a vertical circle under gravity is an archetype of non-uniform circular motion because the tangential component of gravity causes the speed $v$ to vary continuously with height.

Let a small body of mass $m$ tied to an inextensible light string of length $R$ be projected horizontally with velocity $u$ at the lowest point $L$. Let $v$ be its speed when the string makes an angle $\theta$ with the downward vertical.

Height ascended above the lowest point:

Applying the Law of Conservation of Mechanical Energy between lowest point $L$ and position $\theta$:

At position $\theta$, the forces acting on the mass are:

• String tension $T$ directed radially inward toward the center $O$.
• Gravitational force $mg$ directed vertically downward. The component along the string away from the center is $mg\cos\theta$.

The net radial force provides the necessary centripetal acceleration:

Substituting $v^2$ from (Eq. 1):

For the body to execute complete vertical circles without the string slacking, the tension must remain non-negative everywhere, especially at the highest point $H$ where $\theta = 180^\circ$ ($\cos 180^\circ = -1$):

To find the minimum projection velocity $u$ at the lowest point $L$ ($ heta = 0^\circ, h = 2R$), substitute $v_H^2 = gR$ into energy conservation:

At the horizontal position $M$ ($ heta = 90^\circ, h = R$):

At the lowest point $L$ ($ heta = 0^\circ, \cos 0^\circ = 1$):

At the highest point $H$ ($ heta = 180^\circ, \cos 180^\circ = -1$):

From energy conservation, $\frac{1}{2}mu^2 - \frac{1}{2}mv_H^2 = mg(2R) \implies u^2 - v_H^2 = 4gR$.

Subtracting $T_H$ from $T_L$:

The difference in string tension between the lowest and highest points of a vertical circle is always exactly six times the weight of the rotating body, independent of the radius $R$ or initial speed $u$ (provided the loop is completed).

Power ($P$) is the rate at which work is done or energy is transformed per unit time. It is a scalar quantity measuring the rapidity of energy conversion.

• Average Power ($P_{\text{avg}}$): $$P_{\text{avg}} = \frac{\Delta W}{\Delta t}$$
• Instantaneous Power ($P$): The limiting value of average power as the time interval approaches zero: $$P = \lim_{\Delta t \to 0} \frac{\Delta W}{\Delta t} = \frac{dW}{dt}$$ Since $dW = \vec{F} \cdot d\vec{r}$, we obtain the fundamental vector product formulation: $$P = \frac{\vec{F} \cdot d\vec{r}}{dt} = \vec{F} \cdot \left(\frac{d\vec{r}}{dt}\right) = \vec{F} \cdot \vec{v} = F v \cos\theta$$ where $\vec{v}$ is the instantaneous velocity of the body, and $\theta$ is the angle between force $\vec{F}$ and velocity $\vec{v}$.

• SI Unit: Watt ($\text{W}$). $1\,\text{W} = 1\,\text{Joule/second} = 1\,\text{kg}\cdot\text{m}^2\cdot\text{s}^{-3}$.
• Larger Engineering Units:
  • $1\,\text{Kilowatt (kW)} = 10^3\,\text{W}$
  • $1\,\text{Megawatt (MW)} = 10^6\,\text{W}$
  • $1\,\text{Horsepower (hp)} = 746\,\text{Watts}$ (traditional British imperial engineering unit).
• Dimensional Formula: $[M^1 L^2 T^{-3}]$.

Electrical utilities measure energy consumption in kilowatt-hours (kWh), commercially known as the Board of Trade Unit (B.O.T. unit). One kilowatt-hour is the total electrical energy consumed by an appliance of power $1\,\text{kW}$ operating continuously for 1 hour:

Important Trap: Kilowatt-hour (kWh) is a unit of Energy, not Power!

For any motor, machine, or water pump, energy is lost due to frictional dissipation and heating. The efficiency $\eta$ is defined as the dimensionless ratio of output power to input power:

For a water pump lifting water of mass $m$ to height $h$ in time $t$:

A collision is an intense physical interaction between two or more bodies occurring over a relatively small time interval, involving strong mutual forces compared to any external forces present.

Newton defined the Coefficient of Restitution ($e$) as the ratio of the relative velocity of separation to the relative velocity of approach along the common normal (line of impact):

Consider two spherical bodies of masses $m_1$ and $m_2$ moving along a straight line with initial velocities $u_1$ and $u_2$ ($u_1 > u_2$). After head-on elastic collision, let their final velocities be $v_1$ and $v_2$.

1. Conservation of Linear Momentum:

2. Conservation of Kinetic Energy:

Dividing (Eq. 2) by (Eq. 1):

In a 1D elastic collision, the relative velocity of approach equals the relative velocity of separation ($e = 1$).

Solving for final velocities $v_1$ and $v_2$ yields the classical collision formulas:

Special Cases:

• Equal Masses ($m_1 = m_2 = m$): $$v_1 = u_2, \quad v_2 = u_1$$ *Bodies exchange velocities completely upon collision!* If target mass $m_2$ was at rest ($u_2 = 0$), $m_1$ comes to a dead stop ($v_1 = 0$) and $m_2$ moves off with initial speed $u_1$ (100% kinetic energy transfer, as observed in Newtons Cradle).
• Target initially at rest ($u_2 = 0$) and $m_1 \ll m_2$ (Light projectile hits massive wall): $$v_1 \approx -u_1, \quad v_2 \approx 0$$ The light body rebounds with virtually identical speed; massive target remains undisturbed.
• Target initially at rest ($u_2 = 0$) and $m_1 \gg m_2$ (Massive truck hits light ball): $$v_1 \approx u_1, \quad v_2 \approx 2u_1$$ The massive projectile continues forward unabated; light target is propelled forward at double the projectile speed!

In a perfectly inelastic collision ($e = 0$), the colliding bodies coalesce into a single combined mass $(m_1 + m_2)$ moving with a common final velocity $V$:

Loss in kinetic energy during collision (dissipated as thermal, acoustic, and plastic deformation energy):

Since $(u_1 - u_2)^2 > 0$, $\Delta K$ is strictly positive, proving that kinetic energy is always lost in an inelastic collision.