Mechanical Work

Multiply the force by the distance and the cosine of the angle between them: W = F·s·cos(α). If the force pulls exactly along the path (α = 0°, cos = 1), simply W = F·s: a crate pulled with 50 N over 10 m receives W = 500 J. If the force acts at an angle, only its component along the path counts; at α = 60° only half, since cos(60°) = 0.5. If the force is perpendicular to the path (α = 90°), no work is done at all. The unit is the joule: 1 J = 1 N·m. Always check that force is in newtons and distance in metres.

Because the holding force is perpendicular to the path. When carrying horizontally, the force holding the bag points upward (against gravity) while the path runs horizontally; the angle is 90° and cos(90°) = 0, so W = 0. Physical work requires a force component in the direction of motion. That carrying still feels strenuous is biology: muscles consume chemical energy even when merely holding, because their fibres constantly re-tension. You do work in the physical sense only when lifting the bag (lifting work W = m·g·h) or accelerating it. This distinction between physiological effort and physical work is a classic conceptual question.

These are the three most important special cases of W = F·s. Lifting work: raising against gravity, W = m·g·h; for 20 kg raised 2 m that is 20 × 9.81 × 2 ≈ 392 J, stored as potential energy. Acceleration work: making the body faster, W = ½mv² − ½mv₀², which becomes kinetic energy. Friction work: pushing against the friction force, W = F_R·s, which turns into heat and is "lost" to mechanics. In many problems they combine: pushing a crate up a ramp requires lifting work plus friction work. The conservation of energy connects all three; every piece of work done reappears as a form of energy.

Work is the process of energy transfer, energy the stored state; both are measured in joules. If you do work on a body, its energy rises by exactly that amount: lifting work becomes potential energy, acceleration work kinetic energy. The work-energy theorem summarises this: W = ΔE. Conversely, a body with energy can itself do work; water in a reservoir drives turbines as it flows down. A picture helps: energy is the account balance, work the bank transfer. Power then states how fast the transfer happens (P = W/t). Cleanly separating this chain of work, energy and power wins half the marks in mechanics exams.

Because work is defined as the scalar product of force and displacement vectors: W = F⃗·s⃗ = F·s·cos(α). The cosine projects the force onto the path direction; only this share actually pushes the body forward. The transverse component F·sin(α) merely presses it against the surface and contributes nothing. Three limiting cases make this clear: α = 0° gives cos = 1, full effect (pulling along the path); α = 90° gives cos = 0, no work (carrying, centripetal force); α = 180° gives cos = −1, negative work, where the force brakes, like friction acting exactly against the path. Negative work means mechanical energy is being taken from the body.