Force–Displacement Graphs

Check three things before measuring any area. First, the vertical axis must be the component of the force along the motion, not its magnitude: a 50 N rope pull at 30° to a horizontal track belongs on the graph as 50 cos 30° = 43.3 N. Second, the graph belongs to one named force or to the net force. The tension graph gives the tension's work; only the net-force graph gives the change in kinetic energy. Third, the horizontal axis must be position. Width in metres times height in newtons is N m = J, so every grid square has a price: on a grid of 0.5 N by 0.2 m squares each square is worth 0.10 J, and 37 squares under the curve means about 3.7 J. The same shape drawn against time has area in N s — an impulse, a change of momentum — and has no place in an energy calculation.

The work is W = ∫ Fₓ dx from x₁ to x₂, and each strip Fₓ dx carries two signs, one from Fₓ and one from dx. For motion toward +x every dx is positive, so strips above the axis add and strips below subtract. A force of +6.0 N over the first 2.0 m and −3.0 N over the next 3.0 m does +12 J − 9.0 J = +3.0 J; the 21 J of painted area is not a work. Now send the object back from x = 5.0 m to x = 2.0 m through the −3.0 N region. Every dx is negative, so W = (−3.0 N)(2.0 m − 5.0 m) = +9.0 J: the same strip of graph now counts positive, because force and displacement both point toward −x. The rule is that a strip is positive when Fₓ and the displacement share a sign. For a force that depends only on position, the outbound and return trips over the same strip cancel exactly — the graphical reason such a force can have a potential energy.

A graph made of straight segments needs no integration. Cut it at every corner and price each piece with the trapezoid ½(F₁ + F₂)(x₂ − x₁), which already covers rectangles (F₁ = F₂) and triangles (one end zero). A segment that crosses the axis is still one trapezoid for the net work: from +8.0 N at x = 3.0 m to −4.0 N at x = 6.0 m it gives ½(8.0 − 4.0)(3.0) = +6.0 J. That net figure hides the traffic. The crossing sits at x₀ = 3.0 + 8.0 × 3.0/(8.0 + 4.0) = 5.0 m, so the segment is really a +8.0 J triangle (½ × 8.0 × 2.0) followed by a −2.0 J triangle (½ × 4.0 × 1.0). Find x₀ whenever the question is about the most energy delivered or the highest speed, because the running total stops rising exactly there. The furthest point reached is a different landmark: it is where the running total has fallen back by the whole starting kinetic energy, and it lies beyond x₀.

Integrate from a fixed start to a variable end and you get a second graph, W(x) = ∫ Fₓ dx′ from x₁ to x, whose slope is the force: dW/dx = Fₓ. So the work graph can be sketched straight off the force graph — steep where Fₓ is large, flat where Fₓ = 0, falling where Fₓ is negative, with a maximum wherever Fₓ crosses from positive to negative. Plot the net force, add the starting kinetic energy, and the work graph becomes K(x) = K₁ + W(x). A 1.2 kg trolley entering at 5.0 m s⁻¹ carries K₁ = 15 J; if the net-force graph has delivered W = −9.6 J by x = 2.0 m, then K = 5.4 J there and v = √(2 × 5.4/1.2) = 3.0 m s⁻¹. Where K₁ + W(x) reaches zero the object stops, and if the net force there is not zero it turns back, so the graph locates a turning point with no clock and no acceleration.

A force sensor gives readings, not a formula. Join neighbouring points with straight lines and add the trapezoids, W ≈ Σ ½(Fₖ + Fₖ₊₁)(xₖ₊₁ − xₖ), which copes with unequal spacing. A stiffening spring read every 0.10 m gives 0, 1.05, 2.60, 4.65 and 7.20 N out to 0.40 m, so W ≈ 0.05 m × (0 + 2 × 1.05 + 2 × 2.60 + 2 × 4.65 + 7.20) N = 1.19 J. The curve bends upward, every chord lies above it, and the estimate runs high: the quadratic through these points, F = (8.0 N m⁻¹)x + (25 N m⁻²)x², integrates to 1.17 J. Using every other reading gives 1.24 J, an error four times larger for twice the spacing — the trapezoid rule's signature. Calibration errors behave differently: a sensor reading 2% high inflates every strip, and the total, by 2%, and no finer spacing can reveal it.

Everything so far assumes Fₓ has one value at each position. Kinetic friction breaks that: pushing a box out along a floor, Fₓ = −f; bringing it back, Fₓ = +f. On an F–x plot the two trips are two horizontal lines, and the rectangle between them has area 2fL — for f = 4.0 N and L = 1.5 m, 12 J turned into internal energy, which is exactly the magnitude of friction's work over the round trip. A rubber band does the same thing more gently: it pulls harder while stretching than while contracting through the same lengths, and the enclosed loop is the energy it fails to return. A position graph also says nothing about time. Two carts with the same net-force graph gain the same kinetic energy whatever their masses, but they reach different speeds at different moments — which is where power, the next topic, takes over.