Work

In everyday English, “work” means effort. You worked hard at the gym. You pulled an all-nighter. You worked all summer at a coffee shop.

Physics is much more particular. To a physicist, you only do work if you push or pull something and it actually moves and the force has at least some component in the direction of motion. Stand still holding a 50 lb weight overhead for an hour and your muscles will be on fire — but in physics terms, you’ve done zero work on that weight, because nothing moved.

This is one of those sections where intuition will betray you. The MCAT writes great trap questions around it. Lock in the formula, learn the three cases (positive, negative, zero), and the rest is plug-and-chug.

The Work Equation

Work is a scalar, not a vector — it can be positive, negative, or zero, but it has no direction.

The whole reason this equation is more nuanced than “force times distance” is the $\cos\theta$ term. That cosine picks out only the part of the force that lines up with the motion. A force perpendicular to motion does zero work; a force directly opposite to motion does negative work.

The Three Cases: Positive, Negative, Zero

The sign of work tells you whether energy is flowing into or out of the object.

Special Cases Worth Memorizing

• $\theta = 0°$ (force in same direction as motion): $\cos 0° = 1$, so $W = Fd$. Maximum work for a given force and distance. Example: pushing a crate in a straight line behind it.
• $\theta = 180°$ (force opposite to motion): $\cos 180° = -1$, so $W = -Fd$. Negative work. Example: kinetic friction always opposes motion, so friction always does negative work on a sliding object.
• $\theta = 90°$ (force perpendicular to motion): $\cos 90° = 0$, so $W = 0$. Examples: the normal force on a box sliding across the floor; the centripetal force on an object in circular motion.

Work Done by Multiple Forces

When several forces act on an object at once, you have two equally valid options:

1. Calculate work for each force separately, then add them: $W_{net} = W_1 + W_2 + W_3 + \ldots$
2. Find the net force first, then compute: $W_{net} = F_{net} \cdot d \cdot \cos\theta$.

Both give the same answer. Pick whichever is faster for the question in front of you.

Gravity’s Work

Gravity deserves special attention because it shows up in nearly every energy problem. The work done by gravity on an object that moves vertically by height $h$ is:

• $W_{gravity} = -mgh$ if the object moves up (gravity opposes the motion).
• $W_{gravity} = +mgh$ if the object moves down (gravity helps the motion).

If the object moves along a diagonal path (like up a ramp), only the vertical component of displacement matters for gravity’s work. Horizontal motion is perpendicular to gravity, so it contributes nothing. This is why a 10 m walk across a flat floor and a 10 m walk across a higher floor both involve zero gravitational work — but climbing stairs to that higher floor does involve work.

Worked Example

You push a 20 kg box across a frictionless floor with a 30 N force at 60° below horizontal (you’re pushing down on a handle), moving it 5 m. How much work do you do?

• Use $W = Fd\cos\theta$ with $\theta = 60°$.
• $W = 30 \times 5 \times \cos 60° = 30 \times 5 \times 0.5 = 75$ J.

Your full 30 N push covered 5 m, but only $30\cos 60° = 15$ N of it actually pointed forward. The other $30\sin 60° \approx 26$ N pressed straight down into the floor and did nothing useful. That’s why the work is 75 J, not 150 J.