How Do You Calculate Work Done by a Force (W = Fs cos θ)?

For IB Physics SL and HL, calculate the work done by a constant force using $W=Fs\cos\theta$, where $F$ is force, $s$ is displacement, and $\theta$ is the angle between their directions. Work is measured in joules (J).

The Reasoning and Mechanism

Work done is the energy transferred when a force acts through a displacement. Only the force component parallel to the displacement transfers energy:

$$F_{\parallel}=F\cos\theta$$

Multiplying this component by displacement gives:

$$W=F_{\parallel}s=Fs\cos\theta$$

This equation applies directly to a constant force acting during a straight-line displacement. Work is a scalar quantity, although its value may be positive, zero, or negative. Its sign shows whether the force transfers energy to or from the object.

For example, a $50\,\text{N}$ force pulls an object through $4.0\,\text{m}$, acting at $30^\circ$ to the displacement:

$$W=(50)(4.0)\cos 30^\circ=1.7\times10^2\,\text{J}$$

Therefore, the force transfers approximately $170\,\text{J}$ of energy to the object. Equivalently, its parallel component is $50\cos30^\circ=43\,\text{N}$, so $W=(43)(4.0)\approx170\,\text{J}$.

A common misconception is that every force acting on a moving object does work. A perpendicular force, such as the centripetal force in uniform circular motion, does zero work because $\cos90^\circ=0$; it changes the direction of velocity but not the speed.

Exam Technique

This is syllabus topic A.3 Work, energy and power for SL and HL. For a Calculate question, show the equation, substitution, angle, and final value with units. Ensure $\theta$ is the angle between the force and displacement, not automatically the angle from the horizontal.