Physics · Class 11

Work under a force curve

Only the component parallel to displacement contributes to work. A constant force gives a rectangular area. A linearly varying force adds a triangular contribution. Area below zero subtracts from work; a force perpendicular to displacement does no work.

Open the interactive experiment →

Count positive and negative force area.

இடப்பெயர்ச்சிக்கு இணையான விசைக் கூறு மட்டுமே வேலை செய்கிறது. விசை–இடப்பெயர்ச்சி வரைபடத்தின் குறியுடைய பரப்பளவே வேலை.

Key notes, relationships and model limits

Key relationship

W = ∫₀ˣ (F₀ + ks) cosθ ds = (F₀x + ½kx²) cosθ; constant force: W = Fx cosθ.

Remember

Common trap

Force-time area is impulse; force-distance area is work. Their units differ.

Model limits

One straight displacement with a force magnitude coefficient linear in x and fixed direction. Negative coefficient reverses the chosen force direction. This is a work integral, not a solved trajectory; playback scans displacement.

What you will learn

Calculate work from a force-distance curve and interpret angle and sign.

Try the experiment

Try constant force, then a curve crossing zero. Scrub displacement and compare the signed area with the work readout.

Think of it this way

A varying push contributes in small pieces, like adding many narrow strips of area.

Worked example

For F(x) = 5 + 2x N over 5 m, W = 5×5 + ½×2×25 = 50 J. At 60° the same force profile contributes 25 J.

Why this matters

The area method handles force variation without replacing the force by an arbitrary final value.

Before you start

Vector components · velocity · SI units

Check your understanding

A constant force is perpendicular to displacement. Its work is…

Fd
Zero
−Fd

Read the explanation

W = Fd cos90° = 0 because the parallel component vanishes.

Course connection: CBSE XI Chapter 5; NEET/Main Unit 4; JEE Advanced Mechanics. Apply each exam’s stated assumptions.