Topic 3.2 Work: calculate the work done by a force through W = Fd cos(theta), connect net work to the change in kinetic energy, and read work as the area under a force-displacement graph.

What this topic is asking

The College Board (Topic 3.2) wants you to calculate the work done by a force as $W = Fd\cos\theta$, to recognize that work transfers energy to or from an object, to apply the work-energy theorem ($W_{net} = \Delta K$), and to read work as the area under a force-displacement graph. Work is the bridge between forces (Unit 2) and energy (Unit 3): it is how a force changes an object's energy.

The work formula

The factor $\cos\theta$ picks out the component of the force that lies along the direction of motion: only that component transfers energy. A force perpendicular to the motion ($\theta = 90^\circ$) does no work, which is why the normal force on a sliding block and the centripetal force in uniform circular motion do zero work.

Positive, negative, and zero work

Reading the sign correctly is essential. When you push a box forward, your force does positive work and adds kinetic energy. Friction acts backward, so it does negative work and removes kinetic energy. Gravity does positive work on a falling object and negative work on a rising one. The total, the net work, is what changes the kinetic energy.

The work-energy theorem

The work-energy theorem ties work directly to kinetic energy:

The net work done by all forces on an object equals its change in kinetic energy. This is one of the most powerful tools in the course because it lets you find a final speed without tracking the acceleration through time: you sum the work done by every force and set it equal to $\Delta K$. If the net work is positive the object speeds up; if negative, it slows down; if zero, its speed is unchanged.

Work as the area under a graph

When a force varies with position, you cannot just multiply force by distance. Instead, work is the area under the force-versus-displacement graph. For a constant force this area is a rectangle ($W = Fd$); for a linearly varying force, such as a spring, it is a triangle. This graphical view is exactly how the energy stored in a spring, $\tfrac{1}{2}kx^2$, is derived: the force $kx$ grows linearly with stretch, so the area under the line from $0$ to $x$ is a triangle of area $\tfrac{1}{2}(kx)(x) = \tfrac{1}{2}kx^2$. The exam frequently gives a force-displacement graph and asks for the work as the area, so practice computing areas of rectangles, triangles and trapezoids under such curves. This connects the spring force from Topic 2.8 to the spring potential energy in Topic 3.3, and shows why work is the unifying idea: every change in energy in this unit can be traced back to a force acting through a displacement, whether the force is constant or varying.

Try this

Q1. A $25$ N force pushes a crate $3.0$ m in the direction of the force. Calculate the work done. [2 points]

• Cue. $W = Fd\cos 0^\circ = (25)(3.0)(1) = 75$ J.

Q2. A net work of $200$ J is done on a $4.0$ kg object initially at rest. Calculate its final speed. [2 points]

• Cue. $W_{net} = \tfrac{1}{2}mv_f^2$, so $v_f = \sqrt{2(200)/4.0} = \sqrt{100} = 10$ m/s.