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Work Done by a Constant Force

Work is the dot product of force and displacement. Only the component of force along displacement counts.
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Derivation

What work means in physics

In everyday language, "work" means any physical or mental effort. In physics, work has a precise definition that is quite different.

Work is done on a body only when a force causes displacement in the direction of the force.

If you push a wall with all your strength and it does not move — no work is done, however exhausted you feel. If you carry a heavy bag horizontally — the force (upward, supporting the bag) is perpendicular to the displacement (horizontal) — no work is done by the carrying force.

The formula

For a constant force F\vec{F} acting on a body that undergoes displacement d\vec{d}:

W=Fd=FdcosθW = \vec{F} \cdot \vec{d} = Fd\cos\theta

where θ\theta is the angle between the force vector and the displacement vector.

Why cosθ\cos\theta

Only the component of force along the displacement does work. The perpendicular component has no effect on the motion along the displacement direction.

The component of F\vec{F} along d\vec{d} is FcosθF\cos\theta.

Work = (component of force along displacement) × displacement = Fcosθ×d=FdcosθF\cos\theta \times d = Fd\cos\theta.

Sign of work

θ\thetacosθ\cos\thetaWorkMeaning
0°+1+1+Fd+FdForce and displacement in same direction — positive work
90°90°0000Force perpendicular to displacement — zero work
180°180°1-1Fd-FdForce opposes displacement — negative work

Positive work: force aids the motion (engine driving a car)

Zero work: force perpendicular to motion (normal force on a moving body, centripetal force on circular motion)

Negative work: force opposes motion (friction on a sliding body, brakes on a car)

Units

[W]=[F][d]=Nm=J (Joule)[W] = [F][d] = \text{N} \cdot \text{m} = \text{J (Joule)}

One Joule is the work done when a force of 1 N moves a body through 1 m in the direction of the force.

Examples

Pulling a suitcase: Force F=50F = 50 N at angle θ=30°\theta = 30° to horizontal, displacement d=10d = 10 m:

W=50×10×cos30°=500×32433 JW = 50 \times 10 \times \cos30° = 500 \times \frac{\sqrt{3}}{2} \approx 433 \text{ J}

Lifting a box: Force = mgmg upward, displacement = hh upward, θ=0°\theta = 0°:

W=mghcos0°=mghW = mgh\cos0° = mgh

Friction: Friction force ff backward, displacement dd forward, θ=180°\theta = 180°:

Wfriction=fdcos180°=fdW_{friction} = fd\cos180° = -fd

Friction always does negative work on a sliding body.

Work is a scalar

Although force and displacement are vectors, their dot product — work — is a scalar. Work has no direction, only magnitude and sign.

Note
Work depends on displacement, not distance. If a body moves in a circle and returns to its starting point, the net displacement is zero — the net work done by any constant force is zero, regardless of how far the body travelled.