Work and the work-energy theorem

In everyday speech, holding a heavy bag for an hour is hard work. In physics it is no work at all, because the bag does not move. Work is done only when a force moves something.

In Class 9 you met work = force × distance. That is only true when the force points exactly along the motion. In general, only the part of the force along the displacement counts:

• W = Fd cos θ, where θ is the angle between the force and the displacement.
• In vector language, W = F·d, the scalar (dot) product: A·B = AB cos θ = AₓBₓ + A_yB_y + A_zB_z. It is a number, not a vector.

The angle decides the sign:

• θ less than 90°: positive work. Gravity on a falling stone.
• θ more than 90°: negative work. Friction on a sliding box; gravity on a ball going up.
• θ = 90°, or no displacement, or no force: zero work. The string of a stone whirled in a circle always pulls at 90° to the motion, so it does no work.

Example with vectors: a force F = 3î – 2ĵ + 4k̂ N moves a body from the point (–2, 1, 3) m to (3, 6, –2) m. The displacement is d = 5î + 5ĵ – 5k̂ m, so W = 3 × 5 + (–2) × 5 + 4 × (–5) = 15 – 10 – 20 = –15 J.

The unit of work (and of energy) is the joule: 1 J = 1 N m. In CGS units, 1 erg = 1 dyne cm, and 1 J = 10⁷ erg.

The work–energy theorem: the change in a body's kinetic energy equals the work done on it by the net force, K_f – K_i = W. It comes straight from v² = u² + 2as with a = F/m. If the speed does not change, the net force did zero work.

Equal and opposite forces need not do equal and opposite work. When a cycle skids to a stop, the road's friction does negative work on the cycle, but the cycle's push on the road does no work, because the road does not move.