Lessons · energy

Work as area: a variable force

Regime 2 — calculus does what algebra cannot. The integrand isn't constant, so no single algebra product gives the answer; the accumulated quantity is the area under the curve — and SymPy proves that area is exactly the closed-form result.

Regime 2 · calculus does more Math machine-derived & checked by SymPy 4 modeling assumptions (author-asserted)

A 2 kg block on a frictionless track starts at rest. A force that grows with distance, $F(x) = b\,x$ (here $b = 8$ N/m), pushes it for $d = 1.5$ m. How much work is done — and where does $\tfrac12 m v^2$ actually come from? Watch the work accumulate as the area under the force curve.

Work done

W = \tfrac12 b d^2

(the shaded area)

9 J

Kinetic energy gained

\tfrac12 m v^2

9 J

Final speed

v = \sqrt{b/m}\,d

3 m/s

Maximum force

F_{\max} = b\,d

12 N

Work is the area under the force–displacement curve

W = \int F\,dx

Work is defined as the integral of force over displacement. For a constant force this integral is a rectangle (and gives $W = F d$ ); for any force it is the area under the $F$ – $x$ curve. The algebra formula is the special case where the area is a rectangle.

check $\tfrac{d}{dx}W = F$ ; check $W = \int_0^x F\,du$ ; back-substitute into $\tfrac12 m v^2 = W$ ; collapse the constant-force case to $F\,d$

✓ The work's slope is the force: $W'(x) = F(x)$ — the area's rate of growth is the curve's height. [structural]
✓ The accumulated work is exactly the area: $W(x) = \int_0^x F\,du$ . [structural]
✓ The kinetic energy equals the work: $\tfrac12 m v^2 = W$ — the memorized $\tfrac12 m v^2$ is the area. [structural]
✓ When the force is constant the integral collapses to $W = F\,d$ — the algebra formula is the area of a rectangle (the quadrature). [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “Work is the maximum force times the distance,

W = F_{\max}\,d

.”

The force is not constant — it climbs from $0$ to $F_{\max} = b\,d$ . Work is the area under the $F$ – $x$ line, a triangle of area $\tfrac12 F_{\max} d$ — exactly half the rectangle $F_{\max} d$ . Drag the cursor: the shaded triangle, not the enclosing rectangle, is the work, and it equals the kinetic energy gained.

Modeling assumptions — author-asserted, disclosed not discharged

Frictionless track: the applied force is the only force doing work along the motion.
The force acts along the direction of motion (one dimension), so $W = \int F\,dx$ with no angle factor.
Point mass; the block starts from rest ( $v_0 = 0$ ), so all the work becomes kinetic energy.
Ideal linear force law $F(x) = b\,x$ (the force is exactly proportional to displacement).

The F–x graph, fully annotated

A static rendering (Matplotlib): the shaded area under F is the accumulated integral W, and the slope of W is F. The interactive version with a draggable cursor is in the Graph tab above.

Formulas used

Hover a formula to preview its reference entry; click to open it in the reference (or the concept graph):