Work: MDCAT Physics notes
Work notes for MDCAT: W = Fd cos θ, the joule, positive, negative and zero work, work by variable forces from area under F–d graphs, and common traps.
Definition of work
Work is done when a force displaces its point of application. For a constant force $\vec F$ causing displacement $\vec d$:
$$W = \vec F\cdot\vec d = Fd\cos\theta$$
where $\theta$ is the angle between force and displacement. Work is a scalar (dot product of two vectors). In simple language, work = effort × distance moved in the direction of the effort.
SI unit: joule (J). One joule is the work done when a force of 1 N displaces its point of application by 1 m in the direction of the force: $1\ \text{J} = 1\ \text{N m}$.
Positive, negative and zero work
| Angle $\theta$ | $\cos\theta$ | Work | Example |
|---|---|---|---|
| $0^\circ$ | 1 | Maximum positive, $W = Fd$ | Pushing a car forward |
| $0^\circ < \theta < 90^\circ$ | positive | Positive | Pulling a trolley by an inclined handle |
| $60^\circ$ | 0.5 | Half of maximum (50%) | — |
| $90^\circ$ | 0 | Zero | Centripetal force, weight on horizontal motion |
| $90^\circ < \theta \le 180^\circ$ | negative | Negative | Friction, braking force |
Cases of zero work
- Centripetal (radial) force in circular motion, uniform or non-uniform: it is always perpendicular to the velocity. So the Moon orbiting the Earth has no work done on it by gravity; a stone whirled around a full circle has zero work done on it.
- Tension in a simple pendulum: the string pull is along the radius, perpendicular to the bob's motion.
- Weight during horizontal motion: a patient in a wheelchair pushed along a level floor; weight is vertical, displacement horizontal.
- Zero displacement: holding a load stationary does no work on it.
Work against gravity
Lifting a mass $m$ through height $h$ at steady speed: $W = mgh$. Lifting 2 kg through 1.5 m: $W = 2 \times 9.8 \times 1.5 = 29.4\ \text{J}$. Measure $h$ from where the load starts: a load raised from the floor to 0.5 m above a 1.75 m tall lifter's head rises 2.25 m.
Work done by a net force
With several forces, find the net force first. Forces of 15 N and 5 N in opposite directions give a net 10 N; moving 5 m along it gives $W = 50\ \text{J}$.
Work done by a variable force
If the force varies, divide the displacement into small intervals $\Delta d$ over which the force is nearly constant, find $F\cos\theta\,\Delta d$ for each, and add:
$$W \approx \sum F_i\cos\theta_i\,\Delta d_i$$
Graphically, work = area under the force–displacement graph. For a spring, $F = kx$ and the area is a triangle, giving $W = \tfrac12 kx^2$. The spring's own force does negative work while it is being stretched and positive work while it relaxes; over a full return to the starting position its net work is zero.
Key formulas
- $W = Fd\cos\theta$
- $W = mgh$ (lifting)
- Spring: $W = \tfrac12 kx^2$
- $W$ = area under $F$–$d$ graph
Common MDCAT traps
- Work by centripetal force is always zero, not maximum or negative.
- Work is 50% of maximum at $60^\circ$, not $45^\circ$ or $30^\circ$.
- Negative work at $180^\circ$ (force opposite displacement); zero, not negative, at $90^\circ$.
- Area under an $F$–$d$ graph is work, not power; the area under an $F$–$t$ graph is impulse.
- A variable force: split the displacement into small steps.
Quick revision
- Work is a scalar: $\vec F\cdot\vec d$.
- $1\ \text{J} = 1\ \text{N} \times 1\ \text{m}$.
- Maximum work at $\theta = 0^\circ$.
- Full circle under a central force: zero work.
- 10 N moving a body 5 m along it: 50 J.