Introduction
Work represents energy transferred to or from a system by mechanical means. It is done when a force acts through a displacement. Understanding work for both constant and variable forces is fundamental to the energy approach in mechanics, providing powerful alternatives to force-based analysis.
Definition of Work
Work W done by a constant force F acting through displacement d is W = F·d = Fd cosθ, where θ is the angle between force and displacement vectors. Work is a scalar quantity measured in Joules (1 J = 1 N·m). Positive work when force has component in direction of motion (0 ≤ θ < 90°); negative work when force opposes motion (90° < θ ≤ 180°); zero work when force is perpendicular to motion (θ = 90°).
Work by Variable Forces
For a force that varies with position F(x), the work is the integral: W = ∫ F(x)dx from x1 to x2. Geometrically, this equals the area under the F-x curve. In three dimensions: W = ∫ F·dr = ∫ (Fxdx + Fydy + Fzdz) along the path. The work depends on the path taken for non-conservative forces, but is path-independent for conservative forces.
\nIntroduction
Work represents energy transferred to or from a system by mechanical means. It is done when a force acts through a displacement. Understanding work for both constant and variable forces is fundamental to the energy approach in mechanics, providing powerful alternatives to force-based analysis.
Definition of Work
Work W done by a constant force F acting through displacement d is W = F·d = Fd cosθ, where θ is the angle between force and displacement vectors. Work is a scalar quantity measured in Joules (1 J = 1 N·m). Positive work when force has component in direction of motion (0 ≤ θ < 90°); negative work when force opposes motion (90° < θ ≤ 180°); zero work when force is perpendicular to motion (θ = 90°).
Work by Variable Forces
For a force that varies with position F(x), the work is the integral: W = ∫ F(x)dx from x1 to x2. Geometrically, this equals the area under the F-x curve. In three dimensions: W = ∫ F·dr = ∫ (Fxdx + Fydy + Fzdz) along the path. The work depends on the path taken for non-conservative forces, but is path-independent for conservative forces.
\nWork by Multiple Forces
When several forces act on an object, the net work equals the sum of work done by each force: W_net = Σ W_i. Alternatively, find net force first: F_net = Σ F_i, then W_net = F_net·d. Both methods give the same result. The net work done on an object equals its change in kinetic energy (Work-Energy Theorem).
Special Cases
Gravity work: W = -mgΔy (depends only on vertical displacement, not path). Spring work: W = 1/2kx₲ - 1/2kx22 (from x1 to x2, depends only on endpoints). Friction work: W = -f_k·d (always negative, depends on path length). Normal force does no work on a surface (perpendicular to motion). Tension in pendulum does no work (always perpendicular to arc).
Units and Dimensions
SI unit of work and energy is the Joule: 1 J = 1 N·m = 1 kg·m2/s2. Other units: erg (CGS, 1 erg = 10^-7 J), electron-volt (1 eV ≈ 1.6×109 J), kilowatt-hour (1 kWh = 3.6×10^6 J). Dimensions: [Work] = [Energy] = [ML2T⻲]. Work and energy have the same dimensions and units, reflecting their equivalence as different manifestations of the same physical quantity.
\nSolved Example: Pushing a Crate
A worker pushes a 50 kg crate 10 m across a floor with a force of 200 N at 30° below horizontal. Find work done by worker and by gravity. Solution: Work by worker: W = Fd cosθ = 200 × 10 × cos(30°) = 2000 × 0.866 = 1732 J. Work by gravity: Gravity acts vertically down, displacement is horizontal, so θ = 90°. W_gravity = mgd cos(90°) = 0 J. The vertical component of applied force (200sin30° = 100 N downward) increases normal force but does no work since there is no vertical displacement. Worker does 1732 J of work, all transferred against friction (not calculated here) or to kinetic energy if crate accelerates.
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