Variable Forces

The equation $W = Fd\cos\theta$ works perfectly when the force is constant. But what if the force changes as the object moves?

Stretch a rubber band. The first centimeter is easy. The next centimeter is harder. By the time you’ve pulled it 10 cm, you’re really straining. The force isn’t a single number — it’s a different value at every position. You can’t just multiply “the force” by “the distance,” because there’s no single “the force.”

This is where the graphical method comes in. It works for any force — constant, increasing, decreasing, weird-shaped — and the math never goes beyond basic geometry.

The Graphical Method

When force changes with position, work equals the area under the force-vs-displacement (F vs. x) graph.

This is the MCAT-level approach: no calculus required. The MCAT picks problems where the graph is made of straight lines, so the area is always a combination of rectangles, triangles, and trapezoids — shapes you’ve known the area formulas for since middle school.

Common Graph Shapes and Their Areas

Spring Work: The Most Important Triangle

A spring exerts a force proportional to how far it’s stretched or compressed: $F = kx$ (we’ll cover the negative sign and Hooke’s law in §2.5). The force starts at zero when the spring is at its natural length and grows linearly with displacement, so the F vs. x graph is a straight line through the origin.

The area under that line is a triangle:

Notice the $x^2$. Stretching a spring twice as far takes four times as much work, not twice as much. That’s why pulling back a bow string the last few inches is so much harder than the first few.

Reading Complex Graphs

Some MCAT problems give you a piecewise graph — force constant for a while, then changing in a straight line, then maybe constant again. The recipe: break the graph into simple shapes, find each area, and add them up.

For example, if a graph shows a constant 10 N for the first 3 m (rectangle), then a force linearly decreasing from 10 N to 0 N over the next 2 m (triangle):

• Rectangle: $10 \text{ N} \times 3 \text{ m} = 30$ J.
• Triangle: $\tfrac{1}{2} \times 2 \text{ m} \times 10 \text{ N} = 10$ J.
• Total: $30 + 10 = 40$ J.

That’s the entire technique. Identify shapes, sum areas, done.

Positive and Negative Areas

If the force dips below the x-axis (negative force) over some interval, that region contributes negative work — the force was opposing motion in that stretch.

A common scenario: a spring being released from compression. While you compress it, you do positive work on the spring. When you let go and the spring pushes your hand back, the spring does positive work on your hand, and your hand (now resisting) does negative work on the spring (because hand and spring move in the same direction but the spring’s force on the hand is opposite to the hand’s force on the spring — third law again).

When in doubt, always specify which force you’re calculating work for, and on which object.

Worked Example

A bow archer draws back a bowstring that requires linearly increasing force, from 0 N at rest to 200 N at full draw (a draw distance of 0.5 m). How much work does the archer do on the string?

• F vs. x graph: triangle, base = 0.5 m, height = 200 N.
• $W = \tfrac{1}{2}(0.5)(200) = 50$ J.

So 50 J of energy is now stored in the bent bow, ready to fly into the arrow when released. (Roughly the same energy as a small car moving at 1 m/s — modest, but enough to send a light arrow at 60+ m/s.)