Chapter 5 / 5.3 Work

Work, Energy, and Power Notes

The change in kinetic energy of an object equals the work done by the net force acting on it.
Formula: \[ K_f – K_i = W \] where \( K_f = \frac{1}{2}mv^2 \) (final kinetic energy), \( K_i = \frac{1}{2}mu^2 \) (initial kinetic energy), and \( W \) is the work done.
Example: A raindrop falling from 1 km height.
- Work done by gravity: \( W_g = mgh = 10^{-3} \times 10 \times 10^3 = 10 \, \text{J} \).
- Work done by resistive force: \( W_r = \Delta K – W_g = 1.25 \, \text{J} – 10 \, \text{J} = -8.75 \, \text{J} \).

Work is done when a force causes displacement.
Formula: \[ W = F \cdot d \cos \theta \] where \( F \) = force, \( d \) = displacement, and \( \theta \) = angle between force and displacement.
Work is zero if:
- No displacement (e.g., pushing a wall).
- Force is zero (e.g., block sliding on a frictionless table).
- Force and displacement are perpendicular (\( \theta = 90^\circ \), e.g., moon orbiting Earth).

Positive work: Force and displacement in the same direction (\( 0^\circ \leq \theta < 90^\circ \)).
Negative work: Force opposes displacement (\( 90^\circ < \theta \leq 180^\circ \)). Example: Friction slowing a cyclist.
Example: Cyclist skidding 10 m with 200 N friction. Work by road: \[ W = 200 \times 10 \times \cos 180^\circ = -2000 \, \text{J} \]

SI unit: joule (J).
Other units: erg (\( 10^{-7} \, \text{J} \)), electron volt (\( 1.6 \times 10^{-19} \, \text{J} \)), kilowatt-hour (\( 3.6 \times 10^6 \, \text{J} \)).

🔹 Work-Energy Theorem: Relate net work to kinetic energy change.
🔹 Calculate work done by forces (gravity, friction) using \( W = Fd \cos \theta \).
🔹 Understand when work is zero (e.g., perpendicular force and motion).
🔹 Interpret negative work (e.g., resistive forces).