Work–Kinetic Energy Theorem

Newton's second law holds instant by instant: Fₙₑₜ = m dv/dt. To trade time for position, use the chain rule, dv/dt = (dv/dx)(dx/dt) = v dv/dx. Then Fₙₑₜ = mv dv/dx, and since d(½mv²)/dx = mv dv/dx, the right side is the rate at which ½mv² changes with position. Integrate from xᵢ to x_f: ∫ Fₙₑₜ dx = ½mv_f² − ½mvᵢ². The left side is the net work Wₙₑₜ and the right side is the change in kinetic energy ΔK. Nothing required Fₙₑₜ to be constant; a spring, a fading push or a position-dependent drag all fit. In two or three dimensions, dot Newton's law with the velocity: Fₙₑₜ ⋅ v = m (dv/dt) ⋅ v = d(½m v ⋅ v)/dt, so Fₙₑₜ ⋅ dr = dK. The dot product discards the part of the net force perpendicular to the motion, which can bend the path but never change the speed.

A 12.0 kg crate slides 5.00 m down a ramp inclined at 30.0°, with μₖ = 0.250. Three forces act, and each does its own work. Gravity: the crate drops (5.00 m) sin 30.0° = 2.50 m, so W_g = mgh = (12.0 kg)(9.81 m s⁻²)(2.50 m) = +294 J. Normal force: perpendicular to every step, so W_N = 0. Friction: N = mg cos 30.0° = 102 N, so f = μₖ N = 25.5 N and W_f = −(25.5 N)(5.00 m) = −127 J. The net work is the signed sum, Wₙₑₜ = 294 J + 0 − 127 J = +167 J. The other route resolves the forces first: along the ramp Fₙₑₜ = mg sin 30.0° − f = 58.9 N − 25.5 N = 33.4 N, and (33.4 N)(5.00 m) = 167 J again. For a particle the two routes always agree, because every force acts through the same displacement. Drop the signs and you get about 420 J, a number with no meaning. Notice too that the normal force did no work, yet it set the size of friction.

Solve the theorem for the final speed: v_f = √(vᵢ² + 2Wₙₑₜ/m). The square root makes speed changes lopsided. Give a 1200 kg car 60.0 kJ of net work from rest and v_f = √(2 × 60 000 J ÷ 1200 kg) = 10.0 m s⁻¹. Give the same car the same 60.0 kJ when it is already doing 20.0 m s⁻¹, and v_f = √(400 + 100) m s⁻¹ = 22.4 m s⁻¹, a gain of only 2.4 m s⁻¹. Getting from 10.0 to 20.0 m s⁻¹ needs K to rise from 60.0 kJ to 240 kJ, three times the work the first 10.0 m s⁻¹ cost. The sign carries information too: positive net work speeds the object up, negative slows it, zero leaves the speed alone. And if vᵢ² + 2Wₙₑₜ/m comes out negative, no real speed exists: the object stopped before completing the displacement you assumed, so that displacement never happened.

When the forces change partway through — a push released, a slope running onto a rough floor — treat each stage with its own forces and let the kinetic energy at the joint carry over. A box slides from rest down a smooth ramp that drops h = 1.20 m, then onto a level floor with μₖ = 0.400. Stage 1: only gravity does work, since the normal force, even around a smooth bend, is perpendicular to the motion. So ½mv² = mgh and v = √(2gh) = √(2 × 9.81 m s⁻² × 1.20 m) = 4.85 m s⁻¹ at the bottom. Stage 2: friction μₖ mg does −μₖ mg d, and the box stops when that cancels the kinetic energy handed over. Add the two stages and the joint's kinetic energy appears once with each sign and cancels, leaving mgh − μₖ mg d = 0 − 0, so d = h/μₖ = 3.00 m, with the mass gone too. Split the motion when you need the joint speed, or when a stage ends on a condition such as the floor running out; take the whole trip when only the ends matter.

Wₙₑₜ = ΔK is one scalar equation, so it delivers one unknown, and never a direction or a duration. A satellite in a circular orbit has a large net force and a velocity that turns continuously, yet Wₙₑₜ = 0 and its speed never changes, because gravity stays perpendicular to every step. Time is lost the same way. Push a 2.00 kg puck from rest across frictionless ice with 4.00 N for 2.00 m, or with 8.00 N for 1.00 m: both deliver 8.00 J, and both end at √(2 × 8.00 J ÷ 2.00 kg) = 2.83 m s⁻¹. But the first push lasts 1.41 s and the second 0.707 s, which comes from a = F/m and v = at, not from energy. So pick the tool by the unknown. A speed after a known distance calls for work and energy; a time, a direction or a force at an instant calls for Newton's law and kinematics. Many problems need one after the other.

The derivation used one particle, so every force acted through the same displacement as the mass. A rigid object sliding without rotating qualifies; a system with moving parts does not. A 60.0 kg person jumps from a crouch, raising the centre of mass 0.400 m while the feet press on the floor, and leaves the ground at 2.50 m s⁻¹. The floor pushes on feet that do not move, so it does no work, and gravity does −(60.0 kg)(9.81 m s⁻²)(0.400 m) = −235 J. Yet K grows by ½(60.0 kg)(2.50 m s⁻¹)² = 188 J. At least 423 J came from chemical energy in the muscles, an internal store the particle theorem cannot see. A car accelerating on static friction and two skaters pushing apart are the same story. Newton's law still governs the centre of mass, so (N_avg − mg)(0.400 m) = ΔK gives an average floor force of 1.06 kN, but that product is not energy the floor supplied. Such systems need energy accounting that includes internal energy.