Work by a Constant Force

In physics, the word work has a precise mathematical meaning quite different from everyday use. Pushing hard against a wall that doesn't move is exhausting, but no physics work is done.

Concept

The work done by a constant force $\vec{F}$ as a body undergoes displacement $\vec{s}$ is

$W = \vec{F}\cdot\vec{s} = Fs\cos\theta$

where $\theta$ is the angle between $\vec{F}$ and $\vec{s}$. Work is a scalar.

SI unit: joule, $1 \, \text{J} = 1 \, \text{N}\cdot\text{m}$.

Sign of work:

• $\theta < 90°$: $\cos\theta > 0$, work is positive (force has a component along motion, helping it). Pulling a box forward.
• $\theta = 90°$: work is zero (force is perpendicular to motion). The normal force on a sliding block; centripetal force.
• $\theta > 90°$: work is negative (force opposes motion). Kinetic friction; gravity on a body being lifted.

For multiple forces, the net work equals the sum of works done by each force, which equals the work done by the net force.

Derivation

Work is defined so that the change in kinetic energy equals the net work done — this is the motivation for $W = \vec{F}\cdot\vec{s}$.

For a constant net force $\vec{F}$ acting on a mass $m$ initially with velocity $\vec{v}_i$ over displacement $\vec{s}$, using $v_f^2 = v_i^2 + 2\vec{a}\cdot\vec{s}$ (with $\vec{a}=\vec{F}/m$):

$v_f^2 - v_i^2 = \frac{2}{m}\vec{F}\cdot\vec{s}$

Multiply by $m/2$:

$\tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2 = \vec{F}\cdot\vec{s}$

The left side is $\Delta KE$, so $W = \vec{F}\cdot\vec{s}$ is the natural definition.

Worked Example

A block is pulled along a horizontal floor by a rope at 30° above horizontal, with rope tension 50 N. The block moves 8 m horizontally. Find the work done by the rope.

Solution:

$W = F s\cos\theta = (50)(8)\cos 30° = 400 \times \frac{\sqrt{3}}{2} = 200\sqrt{3} \approx 346.4 \, \text{J}$

Note: the vertical component of tension is perpendicular to displacement and does no work; only the horizontal component contributes.

Common Confusions

• "I pushed hard, I did work." If the body didn't move, $s=0$ and $W=0$ — no physics work, regardless of effort.
• Work depends on the angle. Carrying a bag horizontally at constant height: the upward force you exert is perpendicular to horizontal displacement, so you do zero physics work on the bag (ignoring your body's biological energy use).
• Friction can do positive work. On an accelerating car, static friction on the tyres acts forward and does positive work on the car (within bodies, internal forces matter though).
• Work is a scalar. Don't add work as vectors.

Key Takeaways

• $W = \vec{F}\cdot\vec{s} = Fs\cos\theta$.
• Work is positive, zero, or negative depending on the angle between force and displacement.
• SI unit: joule (1 J = 1 N·m).
• Net work = work done by net force = $\Delta KE$ (work-energy theorem).
• Force perpendicular to displacement does no work.