Work and Energy

Conservative forces:

• Work done depends only on initial and final positions, not on the path taken
• Work done in a round trip (closed path) is always zero
• Examples: gravitational force, spring force, electrostatic force

Non-conservative forces:

• Work done depends on the specific path taken
• Work done in a round trip is generally non-zero (typically negative)
• Examples: friction, air resistance, viscous forces

Tests for conservative forces:

1. Round-trip test: If work equals zero for any closed path, the force is conservative
2. Path-independence test: If work between two points is the same regardless of path
3. Mathematical test: If the force can be expressed as the gradient of a scalar potential function

Note: Only for conservative forces can we define a potential energy function.

Power: The rate at which work is done or energy is transferred.

Derivation: $P = \frac{dW}{dt} = \frac{d(\vec{F} \cdot \vec{r})}{dt} = \vec{F} \cdot \frac{d\vec{r}}{dt} = \vec{F} \cdot \vec{v}$

For constant force:

• $P = \vec{F} \cdot \vec{v} = Fv\cos\theta$
• Where $\theta$ is angle between force and velocity

For variable force:

• $P = \vec{F}(t) \cdot \vec{v}(t)$
• Must be evaluated at each instant

Units:

• SI unit: watt (W) = joule/second (J/s)
• Common non-SI unit: horsepower (hp), where 1 hp = 746 W

Example: A car of mass 1000 kg accelerating at 2 m/s² while traveling at 20 m/s develops power: $P = F \cdot v = ma \cdot v = (1000 \times 2) \times 20 = 40,000$ W = 40 kW ≈ 53.6 hp

The kinetic energy of a system of particles equals the sum of the kinetic energies of all individual particles:

$K = \sum_{i} \frac{1}{2} m_i v_i^2$

Where:

• $K$ is the total kinetic energy of the system
• $m_i$ is the mass of each individual particle
• $v_i$ is the velocity of each individual particle

This means:

• Each particle contributes independently to the total kinetic energy
• The system's kinetic energy depends on both the masses and velocities of all components
• Unlike momentum, kinetic energies always add as scalar quantities (not vectors)

For a simple pendulum of length l:

v₁ = √(v₀² - 2gl(1-cosθ))

This follows from conservation of mechanical energy:

• At the lowest point: energy is primarily kinetic (½mv₀²)
• At angle θ: energy is partly kinetic (½mv₁²) and partly potential (mgl(1-cosθ))
• Total energy remains constant: ½mv₀² = ½mv₁² + mgl(1-cosθ)

Example: If a pendulum has v₀ = 3 m/s at its lowest point and l = 0.5m, at θ = 60° its speed would be v₁ = √(9 - 2(9.8)(0.5)(1-0.5)) = 2 m/s

The maximum angle θₘₐₓ occurs when all kinetic energy converts to potential energy (v₁ = 0):

cosθₘₐₓ = 1 - v²/(2gl)

Derivation:

• At maximum angle: ½mv² = mgl(1-cosθₘₐₓ)
• Simplifying: v² = 2gl(1-cosθₘₐₓ)
• Solving for cosθₘₐₓ: cosθₘₐₓ = 1 - v²/(2gl)

Note: For this equation to be valid, v²/(2gl) must be ≤ 2, otherwise the pendulum would complete a full circle.

Example: With l = 0.5m, g = 9.8m/s² and v = 2m/s, θₘₐₓ would be cos⁻¹(1 - 2²/(2×9.8×0.5)) = cos⁻¹(0.59) = 54°

The elastic potential energy of a spring increases by:

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

Where:

• $k$ is the spring constant (stiffness)
• $x$ is the displacement from natural length (elongation or compression)

Key points:

• Potential energy is zero at natural length
• Energy increases quadratically with displacement
• Same formula applies for both compression and elongation
• This stored energy can be converted to kinetic energy when released
• The force required to stretch/compress increases linearly: $F = kx$

Example: A spring with constant 50 N/m stretched 0.2m stores 1 joule of potential energy: $\frac{1}{2}(50)(0.2^2) = 1J$

When a spring is stretched by distance $x$:

Forces:

• Each end experiences force $F = kx$ in opposite directions
• These forces follow Newton's Third Law
• Left end experiences force $kx$ toward the right
• Right end experiences force $kx$ toward the left

Work and Energy Relationship:

• External work done to stretch spring: $W = \int_0^x kx\,dx = \frac{1}{2}kx^2$
• This work is stored as potential energy
• If one end is fixed, only the force at the moving end does work
• The force increases linearly as displacement increases

Energy Conversion:

• When released, this potential energy converts to kinetic energy
• For a mass attached to a spring: $\frac{1}{2}kx^2 = \frac{1}{2}mv^2$ (when fully converted)

The descent height h relates to the initial velocity v of the mass by: $h = v\sqrt{m/k}$

Where:

• m is the mass of the object
• k is the spring constant
• v is the initial velocity

This formula applies when considering a mass-spring system in a gravitational field, though the same result is obtained if gravity is neglected and the equilibrium position is taken as the natural length.

Starting with conservation of energy:

1. Initial state: $\frac{1}{2}mv^2 + \frac{m^2g^2}{2k}$ (kinetic + initial spring potential)

2. At momentary stop: $\frac{1}{2}k(mg/k + h)^2 - mgh$ (spring potential + gravitational potential)

3. Setting them equal: $\frac{1}{2}mv^2 + \frac{m^2g^2}{2k} = \frac{1}{2}k(mg/k + h)^2 - mgh$

4. Solving for h: $h = v\sqrt{m/k}$

This represents the maximum displacement from equilibrium during an oscillation.

Significance of $h = v\sqrt{m/k}$:

• Represents the maximum displacement where a body with initial velocity v momentarily stops
• Provides direct relationship between initial kinetic energy and maximum potential energy
• Shows how energy parameters (m, k, v) determine the turning point in oscillatory motion

Energy conservation connection:

• Initial energy is $\frac{1}{2}mv^2$ (kinetic) plus any initial potential energy
• At maximum displacement h, all kinetic energy has been converted to potential energy
• The equation is derived directly from equating initial and final energy states
• Demonstrates the predictive power of conservation principles in oscillatory systems

This relationship works in both horizontal and vertical systems (with appropriate reference points).