Energy Principles

Work and the Scalar Product

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

$$W = \mathbf{F} \cdot \mathbf{d} = Fd \cos \theta$$

Determines the energy transferred to an object by a force acting over a distance.

$W$=Work done(Joules (J))

$F$=Magnitude of force(Newtons (N))

$d$=Displacement magnitude(meters (m))

$\theta$=Angle between Force and Displacement(degrees)

$\theta = 0°$

→

$W = Fd$ (Maximum positive work — force along motion)

$\theta = 90°$

→

$W = 0$ (No work done — force perpendicular to motion)

$\theta = 180°$

→

$W = -Fd$ (Maximum negative work — force opposes motion)

Scalar Nature: Work is a scalar quantity even though it is derived from two vectors — the dot product always yields a number, not a direction.

Graphical Representation: The area under a Force-Displacement ($F \cos \theta$ vs $d$) graph equals the total work done.

Variable Forces: For changing forces, work equals the area under the $F\cos\theta$ vs $d$ curve: $W = \sum F_i \cos\theta_i \, \Delta d_i$.

Dimensional Check: $[W] = [F][d] = \text{kg} \cdot \text{m/s}^2 \cdot \text{m} = \text{kg} \cdot \text{m}^2/\text{s}^2 = \text{J}$.

Work Done by Gravity and Conservative Forces

Work done by gravity on an object depends only on the vertical height change, not the path taken. This path-independence defines gravity as a conservative force.

$$W_{gravity} = -mgh$$

Work done by gravity when an object is raised by height $h$ (negative because gravity opposes upward displacement).

$m$=Mass of object(kg)

$g$=Gravitational acceleration(9.8 m/s²)

$h$=Vertical height change(m)

Object moves downward

→

$W_{gravity} = +mgh$ (gravity does positive work)

Closed path (return to start)

→

$W_{gravity} = 0$ (net height change is zero)

Path Independence: Whether an object goes straight up, along a ramp, or along a curved path — the work by gravity depends only on $\Delta h$.

Conservative Forces: Gravity, elastic spring force, and electric force are conservative — work done over any closed loop is zero.

Non-Conservative Forces: Friction, air resistance, tension, and applied propulsion forces are non-conservative — work depends on the path length.

Key Distinction: For conservative forces you can define a potential energy; for non-conservative forces you cannot.

Kinetic Energy and the Work-Energy Theorem

Kinetic energy is the energy of an object due to its motion, while the work-energy theorem states that net work done on a body equals the change in its $KE$.

$$W_{net} = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$

Relates the total work performed by all forces to the resultant change in the object's speed.

$m$=Mass(kg)

$v_f$=Final velocity(m/s)

$v_i$=Initial velocity(m/s)

$W_{net} > 0$

→

Object speeds up ($KE$ increases)

$W_{net} < 0$

→

Object slows down ($KE$ decreases)

$W_{net} = 0$

→

Speed is unchanged (constant velocity)

Velocity Dominance: $KE \propto v^2$ means doubling speed → 4× the kinetic energy, tripling speed → 9× the kinetic energy.

Net Work: Only the net (resultant) force contributes to the change in kinetic energy — individual forces may do positive or negative work.

Dimensional Check: $[KE] = \text{kg} \cdot (\text{m/s})^2 = \text{kg} \cdot \text{m}^2/\text{s}^2 = \text{J}$.

Potential Energy: Gravitational and Elastic

Gravitational potential energy near Earth's surface is energy stored due to an object's height above a chosen reference level.

$$U_g = mgh$$

Calculates energy stored due to vertical position in a uniform gravitational field.

$m$=Mass(kg)

$g$=Gravitational acceleration(9.8 m/s²)

$h$=Height above reference level(m)

$h = 0$

→

Gravitational P.E. is zero (at reference level)

$h < 0$ (below reference)

→

P.E. is negative — perfectly valid

Reference Level: P.E. is relative — only $\Delta U_g = mg\Delta h$ has physical meaning. Both students in the textbook are correct depending on their chosen reference.

Limitation: $U_g = mgh$ is valid only near Earth's surface where $g$ is approximately constant.

Elastic potential energy is energy stored in a spring (or any elastic object) due to its compression or extension from equilibrium.

$$U_s = \frac{1}{2}kx^2$$

Calculates energy stored in a spring displaced from its natural length.

$k$=Spring constant (stiffness)(N/m)

$x$=Displacement from equilibrium(m)

$x = 0$

→

No stored energy (natural length)

Compression or extension

→

Same $|x|$ stores the same energy — $x^2$ makes sign irrelevant

Restoring Force: Elastic P.E. comes from work done against the spring's restoring force $F = -kx$ (Hooke's Law).

Quadratic Dependence: $U_s \propto x^2$ means doubling the compression stores 4× the energy — not 2×.

Energy Exchange in SHM: In a mass-spring system, elastic PE and KE continuously interchange, but their sum $\frac{1}{2}kx_0^2$ is constant.

Conservation of Mechanical Energy

In an isolated system with only conservative forces, the total mechanical energy (sum of $KE$ and $PE$) remains constant throughout the motion.

$$KE_i + PE_i = KE_f + PE_f$$

Predicts the state of a system by assuming energy is transformed but never created or destroyed.

$KE$=Kinetic Energy(J)

$PE$=Potential Energy (Gravitational or Elastic)(J)

Non-conservative forces present

→

$KE_i + PE_i = KE_f + PE_f + W_{friction}$

Object released from rest at height $h$

→

Speed at bottom: $v = \sqrt{2gh}$ (mass cancels!)

Mass Cancels: For a falling object, $mgh = \frac{1}{2}mv^2$ → $v = \sqrt{2gh}$. The final speed is independent of mass — a feather and a bowling ball fall equally fast in vacuum.

With Friction: $mgh = \frac{1}{2}mv^2 + f \cdot d$ — friction steals energy and converts it to heat.

Percentage Lost: Energy lost to friction as a fraction: $\frac{f \cdot d}{mgh} \times 100\%$.

Interconversion: At any point during free fall: $PE + KE = mgh_{initial}$. As PE decreases, KE increases by exactly the same amount.

Power

Power is the rate of doing work, measuring how quickly energy is transferred or transformed.

$$P = \frac{W}{t} = Fv$$

Measures how fast energy is being transferred or transformed.

$P$=Power(Watts (W) = J/s)

$W$=Work done(Joules (J))

$t$=Time interval(seconds (s))

$v$=Velocity (constant)(m/s)

1 Horsepower

→

≈ 746 Watts

1 kWh

→

$= 3.6 \times 10^6$ J = 3.6 MJ (a unit of energy, not power)

Instantaneous Power: $P = \mathbf{F} \cdot \mathbf{v}$ gives power at any instant when force and velocity may vary.

Practical Application: Climbing stairs — same work ($mgh$) done quickly requires more power than doing it slowly.

Absolute Potential Energy and Escape Velocity

Far from Earth's surface, $g$ varies with distance. The absolute gravitational potential energy accounts for the $1/r$ dependence of the gravitational force.

$$U = -\frac{GMm}{r}$$

Gives the gravitational PE of mass $m$ at distance $r$ from Earth's center, with zero PE defined at infinity.

$G$=Universal gravitational constant($6.674 \times 10^{-11}$ N m²/kg²)

$M$=Mass of Earth($5.97 \times 10^{24}$ kg)

$m$=Mass of object(kg)

$r$=Distance from Earth's center(m)

$r = R$ (Earth's surface)

→

$U_s = -\frac{GMm}{R}$ (most negative PE — deepest in gravitational well)

$r \to \infty$

→

$U \to 0$ (object is free from gravitational influence)

Negative Sign: The negative sign means the object is gravitationally bound to Earth. Work must be done to increase $U$ toward zero.

Why Negative?: Zero PE is at infinity. Any finite $r$ means the object has fallen into the gravitational well, so PE is below zero.

Connection to $mgh$: Near the surface, $-\frac{GMm}{R+h} - (-\frac{GMm}{R}) \approx mgh$ when $h \ll R$.

The escape velocity is the minimum initial speed needed for an object to escape a planet's gravitational field entirely, reaching infinity with zero residual speed.

$$v_{esc} = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$$

Derived by equating initial KE to the absolute PE at the surface: $\frac{1}{2}mv_{esc}^2 = \frac{GMm}{R}$.

$v_{esc}$=Escape velocity(m/s)

$G$=Universal gravitational constant(N m²/kg²)

$M$=Mass of the planet(kg)

$R$=Radius of the planet(m)

Earth

→

$v_{esc} = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} \approx 11.2$ km/s

Moon

→

$v_{esc} \approx 2.4$ km/s (smaller mass and radius)

Mass Independence: $v_{esc}$ does not depend on the mass of the escaping object — a molecule and a rocket need the same escape speed.

Not Launch Speed: Real rockets don't need to reach 11.2 km/s instantaneously — they burn fuel continuously. Escape velocity applies to unpowered projectiles.

Derivation Shortcut: Set $KE = |U_{surface}|$: $\frac{1}{2}mv^2 = \frac{GMm}{R}$, the $m$ cancels immediately.