Work by a Constant Force

Split any force into a part along the object's velocity and a part perpendicular to it. Newton's second law says the parallel part changes the speed and the perpendicular part only bends the path. Work is defined to keep the first and discard the second: for a constant force whose point of application moves through Δr, W = F∥ Δr = F Δr cos θ. Take a rope pulling a crate with 40 N at 30° above the floor while the crate slides 5.0 m. The along-floor part is 40 cos 30° = 34.6 N, so W = 34.6 N × 5.0 m = 173 J. The upward part, 40 sin 30° = 20 N, moves nothing vertically and contributes nothing, although it does press the crate less firmly into the floor. The unit is the joule, 1 J = 1 N⋅m = 1 kg m² s⁻²: the energy transferred when a 1 N force acts through 1 m along its own direction.

The θ in F Δr cos θ is the angle between F and Δr placed tail to tail, and it is often not the angle printed on the diagram. A 2.0 kg parcel slides 3.0 m down a chute inclined at 25°. Its weight points straight down and its displacement points 25° below the horizontal, so the angle between them is 90° − 25° = 65°, and W_g = (19.6 N)(3.0 m) cos 65° = 24.9 J. Using the 25° by mistake gives 53.3 J, more than double. Components avoid the trap. With x horizontal and y up, F = (0, −19.6) N and Δr = (3.0 cos 25°, −3.0 sin 25°) m = (2.72, −1.27) m, so W = FₓΔx + F_yΔy = 0 + (−19.6 N)(−1.27 m) = 24.9 J. Two negative components multiply to positive work, as they must for a parcel moving the way gravity pulls. When vectors arrive as components, use components; when they arrive as a size and a direction, find the tail-to-tail angle first.

Three cases cover every constant force. With θ below 90° the force has a component along the motion and does positive work: gravity on a falling stone, the rope on the crate. With θ above 90° that component points against the motion and the work is negative: kinetic friction on a sliding block, or gravity on a 0.50 kg ball rising 2.0 m, which receives (0.50 kg)(9.81 m s⁻²)(2.0 m) cos 180° = −9.8 J. With θ at exactly 90°, or with no displacement of the point where the force acts, W = 0: the normal force on a block sliding along a level floor, a waiter's upward push on a tray he carries across the room, your grip on a 20 kg suitcase held still. That last one feels like effort, and your muscles do burn chemical energy, but it stays inside you; none of it is transferred to the suitcase. Settling the sign first also checks the arithmetic: a result whose sign contradicts the picture has a wrong angle in it.

F⋅Δr uses the displacement, not the distance travelled, so for a force of fixed size and direction only the start and end points matter. Throw the 0.50 kg ball up 2.0 m and catch it where it left your hand: gravity does −9.8 J on the way up and +9.8 J on the way down, and the whole-trip formula agrees at once, since Δy = 0. Now shove a box 1.5 m across a floor and drag it 1.5 m back against 6.0 N of kinetic friction. Again the net displacement is zero, but friction reverses when the box does, so it is not one constant vector. It does −9.0 J on each leg, −18 J in all, never zero. Whenever a force changes direction part-way, split the motion into legs over which the vector really is constant. And always take Δr at the point where the force is applied: lean on a wall that does not move and your push does no work, however long you keep it up.

Most objects feel several forces at once, and each does its own work over the same displacement. Return to the crate: mass 15 kg, pulled with 40 N at 30° above the floor, sliding 5.0 m, with μₖ = 0.20. The rope does +173 J. The normal force is not mg here, because the rope's 20 N upward part carries some of the weight: N = 147.2 N − 20.0 N = 127.2 N, so friction is f = μₖN = 25.4 N and W_f = −127 J. The weight and the normal force are perpendicular to Δr and do zero. The net work is the sum of the entries, Wₙₑₜ = 173 J − 127 J = 46 J. As a check, find the net force first, 34.6 N − 25.4 N = 9.2 N along the floor, and dot it once: 9.2 N × 5.0 m = 46 J. Notice what the rope's vertical part did: no work itself, yet by lightening the load on the floor it cut friction's work by 20 J. Individual work says what one force did; net work says what all of them did together.

Inertial observers agree on forces but not on displacements, so work depends on the frame. A 60.0 kg passenger stands in a lift rising at a steady 2.00 m s⁻¹. Over 5.00 s the floor pushes up with N = mg = 589 N while the passenger rises 10.0 m, so in the building's frame the floor does +5.89 kJ of work and gravity does −5.89 kJ. In the lift's own frame, inertial too because it moves at constant velocity, the passenger does not move at all, and both forces do zero work. The individual entries disagree between frames, and both observers are right. They agree on the net work, zero, only because the net force is zero. When it is not, even the net work shifts with the frame, by Fₙₑₜ⋅uΔt for a frame moving at velocity u. So never mix frames inside one calculation: choose the frame, measure every displacement in it, and let every entry of the ledger use that one choice.