Work by a Variable Force

Constant-force work is a multiplication, so make the force constant by force of numbers. A hand stretches a spring of k = 200 N m⁻¹ slowly from x = 0.10 m to 0.30 m, pulling with F(x) = 200x. Cut the 0.20 m into four slices of 0.05 m and freeze the force at the start of each: (20 + 30 + 40 + 50) N × 0.05 m = 7.0 J. Freeze it at the end of each instead: (30 + 40 + 50 + 60) N × 0.05 m = 9.0 J. The true work is trapped between the two. With 40 slices the sums are 7.9 J and 8.1 J: ten times as many slices, one tenth of the gap. The limit of the sum is the definite integral, and you never have to take it by brute force. Find any antiderivative of the force, a function A(x) with A′(x) = F(x); then W = A(x₂) − A(x₁). Here A(x) = 100x², so W = 100(0.090 − 0.010) = 8.0 J. For a constant force A(x) = Fx, and the integral hands back the familiar F(x₂ − x₁).

Plot F against x and each slice F(xᵢ)Δx becomes a thin strip, so the integral is the area between the curve and the axis, measured on the axes' own scales in N × m = J rather than in squares of paper. For motion toward +x, strips above the axis add work and strips below subtract it. Let the spring itself act on a block moving from x = −0.10 m to +0.30 m, with F = −200x. The triangle above the axis, 0.10 m wide and 20 N tall, gives +1.0 J; the triangle below, 0.30 m wide and 60 N deep, gives −9.0 J; the net is −8.0 J, matching −½k(0.30² − 0.10²). The part from −0.10 m to +0.10 m cancels itself, which is why this equals the spring's work over 0.10 m to 0.30 m alone. Watch the running total too: W(x) = ∫ F dx from x₁ to x has slope dW/dx = F(x), so the accumulated work climbs while F is positive, reaches its peak of +1.0 J at x = 0 where the force changes sign, and falls from there.

On a line, F and dx are both signed numbers. Motion toward −x means every dx is negative, and the integral records that by running from a larger x₁ to a smaller x₂, with no cos θ required. Release a block from x = +0.30 m on the same spring. On the way to x = 0: W = −100(0² − 0.30²) = +9.0 J, positive because the force and the motion both point toward −x. Carrying on to x = −0.30 m: W = −100[(−0.30)² − 0²] = −9.0 J, negative because the force has turned round while the motion has not. One formula, fed the actual start and end positions, gets both legs right. The common slip is to write the force as its magnitude kx and pick a sign by eye: that works until the block crosses x = 0, where the force reverses and the eye usually does not. And because swapping the limits of an integral negates it, retracing a trip under a force that depends on x alone exactly undoes the work.

Three antiderivatives cover most first-year problems. For a spring, F = −kx, A(x) = −½kx², so W = −½k(x₂² − x₁²). For inverse-square gravity along a radius, Fᵣ = −GMm/r², A(r) = GMm/r, so W = GMm(1/r₂ − 1/r₁); with GM = gR², a climb h from Earth's surface gives W = −mgRh/(R + h). For h = 10 km that differs from −mgh by the factor R/(R + h) = 0.9984, a 0.16% correction that justifies constant g near the ground; by a 400 km orbit the gap has grown to about 6%. A power law F = cxⁿ integrates to c(x₂ⁿ⁺¹ − x₁ⁿ⁺¹)/(n + 1) for every n except −1, where the answer is a logarithm: air held at constant temperature in a cylinder pushes its piston with F = C/x, so W = C ln(x₂/x₁). In each case write the force as a signed component first, then integrate, and the sign of the result needs no further thought.

Everything so far assumed one force for each position. Then the work from x₁ to x₂ depends only on x₁ and x₂: an excursion out to x₃ and back adds an integral from x₂ to x₃ plus its exact negative, and any round trip does zero work. Kinetic friction breaks the assumption, because it points against the velocity rather than toward any place. Push a 20 kg crate (μₖ = 0.30) 3.0 m along a floor and back again: friction is −58.9 N on the way out and +58.9 N on the way back, so its force-position graph is a rectangle, not a curve, and the round trip does −58.9 N × 6.0 m = −353 J of work with zero displacement. That is minus the area of the loop, 117.7 N tall and 3.0 m wide. Rubber bands, tendons and shoe foams do the same more gently. They pull harder while being stretched than while relaxing, so the loading curve sits above the unloading one, and the area between them is the energy each cycle turns into heat.

Four quick tests catch most mistakes. Units: F in N times x in m must give J, and GMm/r has units N m² kg⁻² × kg² ÷ m = N m. Bounds: for motion toward +x the work lies between Fₘᵢₙ(x₂ − x₁) and Fₘₐₓ(x₂ − x₁), so the hand's 8.0 J on the spring must sit between 20 N × 0.20 m = 4.0 J and 60 N × 0.20 m = 12 J, and it does. Limits: shrink the displacement and W must approach F(x₁)Δx; shrink the climb and the exact gravity result must approach −mgh. Signs: ask whether the force points with or against the motion over most of the trip, and whether the result agrees. An answer that fails any one of these is wrong before you go hunting for the algebra slip. An answer that passes all four is not proven right, but whatever error remains is no longer a gross one.