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.

1. Work done by the spring force: For displacement from 0 to $x$: $W = \int_0^x -kx \, dx = -\frac{1}{2}kx^2$

   The negative sign indicates work is done against the spring force.

2. Elastic potential energy: $U_{\text{elastic}} = \frac{1}{2}kx^2$

   This represents energy stored in the spring when compressed or extended by distance $x$.

   Note: The potential energy is positive regardless of whether the spring is compressed or extended, but the work done by the spring force is negative when working against it.

Changes in kinetic energy due to external forces:

1. General relationship:

   • Kinetic energy changes when work is done on the particle
   • $\Delta K = W = \int \vec{F} \cdot d\vec{r}$

2. Force angle effect:

   • Maximum energy transfer occurs when force is parallel to velocity
   • Zero energy transfer occurs when force is perpendicular to velocity
   • For any angle θ between force and velocity: $\frac{dK}{dt} = F\cdot v\cos\theta$

3. Special condition: No kinetic energy change occurs when:

   • The resultant force is perpendicular to velocity ($\theta = 90°$)
   • Examples: uniform circular motion, where centripetal force is always perpendicular to velocity
   • In such cases, the speed remains constant while direction changes

4. Mathematical proof:

   • When $\vec{F} \perp \vec{v}$, then $\vec{F} \cdot \vec{v} = 0$
   • Since $\frac{dK}{dt} = \vec{F} \cdot \vec{v}$, the kinetic energy remains constant

The work done equals zero in these transitions because:

• When moving from elongated (x) to compressed (-x), the path must pass through the natural length (0)
• The complete path can be broken into two segments:
  1. Elongated to natural: Work = +½kx² (positive)
  2. Natural to compressed: Work = -½kx² (negative)
• The total work is the sum: +½kx² + (-½kx²) = 0

This illustrates the path-dependent nature of work for non-conservative forces, where the net work in a round trip equals zero. The spring force is conservative, meaning the work done depends only on initial and final positions, not the path taken.

A conservative force is one where the work done in a round trip is zero, while for a non-conservative force like friction, the work done in a round trip is non-zero.

Specifically for friction:

• When a block of mass m is dragged distance l on a surface with friction coefficient μ:
  • The normal force is N = mg
  • The friction force is f = μmg
  • Work done by friction over a distance l is W = -μmgl
  • When dragged back to starting position, work is again W = -μmgl
  • Total work in round trip = -2μmgl (always negative)

This negative work represents energy permanently lost to heat, unlike conservative forces (like gravity or springs) where energy is recoverable.

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: $\frac{1}{2}(50)(0.2^2) = 1J$

To find maximum displacement when a block with initial velocity attaches to a spring:

Apply conservation of mechanical energy:

• Initial energy = Final energy at maximum displacement
• Initial: only kinetic energy $\frac{1}{2}mv^2$ (spring at natural length)
• Final: only potential energy $\frac{1}{2}kx_{max}^2$ (velocity = 0 at maximum displacement)

Therefore: $\frac{1}{2}mv^2 = \frac{1}{2}kx_{max}^2$

Solving for maximum displacement: $x_{max} = v\sqrt{\frac{m}{k}}$

Where:

• $v$ is initial velocity
• $m$ is mass of block
• $k$ is spring constant

Example: A 2kg block moving at 3m/s encounters a spring with constant 12N/m. Maximum displacement will be $3\sqrt{\frac{2}{12}} = 1.22m$

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.

Mass-energy equivalence: $E = mc^2$

Where:

• E is energy
• m is mass
• c is speed of light in vacuum (3×10⁸ m/s)

For an electron:

• Energy equivalent ≈ 8.18×10⁻¹⁴ J
• This is the energy released if an electron is completely converted to energy
• The same amount of energy would be required to create an electron

This relationship demonstrates that mass itself is a form of energy.

This simplification is valid when:

• You're primarily interested in the dynamics relative to the equilibrium position
• The spring is already pre-stretched by gravity at equilibrium
• You redefine the zero point of potential energy at the equilibrium position
• You're analyzing oscillations around the equilibrium

This approach simplifies calculations because:

• The equilibrium position already incorporates gravitational effects
• Deviations from equilibrium follow the same mathematical form as a horizontal spring
• The natural oscillation frequency remains $\omega = \sqrt{k/m}$

Note: This simplification yields the same results as when gravity is explicitly included in the analysis.