Work done by variable force — integration

WHY do we even need integration?

WHAT is the problem? $W=Fd$ assumes $F$ is the same number everywhere along $d$ . If $F$ depends on position, $F(x)$ , that formula is a lie except over a vanishingly small step.

WHY does integration fix it? Because $F(x)$ is approximately constant over an interval $dx$ that is infinitely small. So on that step the work really is a rectangle $F(x)\,dx$ . The integral $\int F(x)\,dx$ is defined as the limit of summing those rectangles — it is literally "area under the $F$ vs $x$ graph."

HOW do we derive it from scratch? (1‑D case, force along motion.)

Figure — Work done by variable force — integration

Worked Examples

Worked example 1 — Stretching a spring (Hooke's law)

A spring obeys $F_{\text{applied}} = kx$ (you must pull with this to balance the spring). Find work to stretch from $0$ to $x_0$ . $W = \int_0^{x_0} kx\,dx$ Why this step? Force grows linearly with stretch, so we cannot use $Fd$ ; we integrate. $W = k\left[\frac{x^2}{2}\right]_0^{x_0} = \frac{1}{2}k x_0^2$ Why this answer makes sense: the average force over the stretch is $\frac{0 + kx_0}{2} = \tfrac12 kx_0$ , and work = average force × distance $= \tfrac12 kx_0 \cdot x_0 = \tfrac12 kx_0^2$ . ✓ Geometrically it's the triangle under the line $F=kx$ .

Common Mistakes

Recall Feynman: explain it to a 12-year-old

Imagine pushing a shopping cart, but the floor gets stickier the farther you go. At the start it's easy, later you push really hard. To find total effort, you can't just multiply one push by the distance — the push keeps changing! So you pretend you walk one tiny step at a time, note how hard you pushed on that step, multiply by the tiny step, and write it down. Do this for every tiny step and add all the little numbers. That giant addition of "push × tiny step" is exactly what mathematicians call an integral, and it gives the total work. It's also the area of the picture you'd draw with "push" going up and "distance" going across.

Flashcards

What is the general definition of work for a variable force?

W = \int_A^B \vec F \cdot d\vec r

, the sum of

\vec F\cdot d\vec r

over the path.

Why can't we use

W=Fd

for a variable force?

Because

F

changes along the path;

Fd

assumes a single constant force value.

How does the integral

\int F\,dx

relate to a graph?

It is the area under the Force vs position curve.

Derive the work to stretch a spring from 0 to

x_0

W=\int_0^{x_0}kx\,dx=\tfrac12 kx_0^2

What does the area under an

F

–

x

graph represent?

The work done by the force.

For

F(x)=3x^2+2x

from

x=1

x=3

, what is

W

[x^3+x^2]_1^3 = 36-2 = 34

When is work done by a force negative?

When the force has a component opposite to displacement (

\cos\theta<0

Why does a perpendicular force (e.g. normal/centripetal) do zero work?

Because

\vec F\cdot d\vec r = F\,dr\cos90^\circ = 0

What is the Riemann-sum origin of

W=\int F\,dx

W=\lim_{\Delta x\to0}\sum F(x_i)\Delta x_i

Connections

Work done by a constant force — the special case $F=$ const where the integral reduces to $Fd$ .
Hooke's law and spring potential energy — direct application giving $\tfrac12 kx^2$ .
Work–Energy Theorem — $\int F\,dx = \Delta KE$ links this to kinetic energy.
Area under curves and the definite integral (Maths) — the mathematical machinery.
Conservative forces and potential energy — when $\int F\,dx$ is path-independent.
Dot product and components of vectors — justifies the $\cos\theta$ factor.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, jab force constant hota hai to work nikalna easy hai: bas $W = F\times d$ , ek rectangle ka area. Lekin real life mein force badalta rehta hai — jaise spring ko jitna zyada khincho, utna zyada force lagta hai. Aise case mein $Fd$ formula jhooth bol deta hai, kyunki force har point pe alag hai.

To trick ye hai: poore path ko bahut chhote-chhote tukdo (slices) mein kaat do, itne chhote ki har slice pe force almost constant lage. Ab har slice ka work ek tiny rectangle ban jata hai — $F(x)\,dx$ . Phir saare tiny rectangles ka area jod do. Ye jodna hi integration hai: $W = \int F(x)\,dx$ . Yaani work = Force vs position graph ke neeche ka area.

Spring ka classic example: $F = kx$ , to stretch karne ka work $\int_0^{x_0} kx\,dx = \tfrac12 k x_0^2$ . Notice karo ki average force $\tfrac12 kx_0$ hai, aur usko distance se multiply karo — wahi answer. Graph pe ye ek triangle ka area hai, simple!

Yaad rakhna do baatein: (1) jaise hi force position ya velocity pe depend kare, turant integral sign uthao, $Fd$ mat lagao. (2) Agar force motion ke opposite hai to work negative hota hai — energy bahar nikal rahi hai. Mantra: "Slice it, push it, sum it."

Work done by variable force — integration

WHY do we even need integration?

Worked Examples

Common Mistakes

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Work, Energy & Power

Connections