3.2 Work

Work is the energy a force transfers into or out of a system as the force acts over a distance, and you find it with $W = F {\parallel} d = Fd\cos\theta$. Work is a scalar that can be positive, negative, or zero, and the work energy theorem ($\Delta K = \sum W i$) ties net work directly to changes in kinetic energy. These ideas connect forces, motion, and energy changes in physical systems.

Why This Matters for the AP Physics 1 Exam

Work is the bridge between forces and energy, so it shows up everywhere in Unit 3 and connects back to the dynamics ideas in Unit 2. Once you can calculate work from a force, an angle, or a graph, you can analyze how a system's kinetic energy changes without tracking every detail of the motion.

This topic gives you strong practice for the free-response question that focuses on building and using mathematical models. You may need to calculate a quantity, use or create a graph, and write a clear explanation that cites physics principles. The work-energy theorem and the area-under-the-curve method are exactly the kinds of tools that question rewards. Work concepts also show up in multiple-choice questions about signs of work, perpendicular forces, and energy transfer.

Key Takeaways

• Work transfers energy into or out of a system: positive work adds energy, negative work removes it, and zero work changes nothing.
• Use $W = Fd\cos\theta$ for a constant force, where $\theta$ is the angle between the force and the displacement of the point where the force acts.
• Only the force component parallel to the displacement does work; a perpendicular component can turn the motion but cannot change kinetic energy.
• The work-energy theorem says net work equals the change in kinetic energy: $\Delta K = \sum W_i$.
• Work equals the area under a graph of $F_{\parallel}$ versus displacement, including negative areas below the axis.
• Conservative forces (like gravity and springs) do path-independent work that is zero around a closed loop; nonconservative forces (like friction and air resistance) do path-dependent work.

Work Done by Forces

When a force acts on a system as it moves over a distance, that force transfers energy into or out of the system. That energy transfer is what we call work, and it can raise or lower the system's energy depending on the direction of the force relative to the motion.

Conservative vs. Nonconservative Forces

Forces split into two groups based on how their work depends on the path.

Conservative forces do work that depends only on the starting and ending configurations, not the route taken:

• If the system returns to its initial configuration, the total work done by a conservative force is zero.
• Potential energy can be associated only with conservative forces.
• Gravity and spring forces are conservative.

Nonconservative forces do work that depends on the actual path:

• Friction and air resistance are the standard examples.
• Their work usually shows up as thermal energy or sound.
• Over a closed path, the work done by a nonconservative force is generally not zero.

This path dependence is why a longer, rougher route loses more energy to friction even when start and end points are the same.

Work Is a Scalar

Work has magnitude but no direction. Its sign tells you which way energy flows.

Positive work:

• The parallel force component points the same way as the displacement.
• Energy is added to the system (example: pushing a cart forward).

Negative work:

• The parallel force component opposes the displacement.
• Energy is removed from the system (example: brakes slowing a car).

Zero work:

• Force and displacement are perpendicular ($\cos 90^\circ = 0$), or
• There is no displacement at all (example: holding a heavy book still).

Work by a Constant Force

For a constant force, work depends on the component of the force parallel to the displacement of the point where the force is applied.

$W = F_{\parallel} d = Fd \cos \theta$

Where:

• $W$ is work in joules (J)
• $F$ is the force in newtons (N)
• $d$ is the displacement of the point where the force acts, in meters (m)
• $\theta$ is the angle between the force and displacement vectors

Only the parallel component changes the system's total energy. When several forces act, the net work is the sum of the work done by each force, which is why you can write it as $W_{net} = \sum W_i = \sum F_{\parallel,i}\, d$.

Force Components and Displacement

Break a force into components relative to the motion:

• Parallel component: $F_\parallel = F\cos\theta$ (does work)
• Perpendicular component: $F_\perp = F\sin\theta$ (does no work)

The parallel component either increases the object's energy (force and displacement same direction) or decreases it (force opposes displacement). The perpendicular component can change the direction of the center-of-mass motion without changing kinetic energy. This is exactly why a normal force on a flat surface, or the tension in a string for circular motion, does no work.

Work-Energy Theorem

The work-energy theorem links net work to the change in kinetic energy:

$\Delta K = \sum_i W_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

Net work done on an object equals its change in kinetic energy. You can apply this to an object, or to a system when the center of mass and the point of application of the force move the same distance. In that case the system can be modeled as an object, and the external work changes only kinetic energy.

This theorem is most useful when you can find net work by adding up individual works or by reading the area under an $F_{\parallel}$-versus-displacement graph.

External Forces and System Configuration

Work depends on the displacement of the point where the force is applied, not automatically the displacement of the system's center of mass. If an external force stretches, compresses, or reshapes a system, the work is still $W = F_{\parallel} d$, where $d$ is the displacement of the point of application.

• If the point of application and the center of mass move the same distance, treat the system as an object, and only kinetic energy changes.
• If parts of the system move different distances, the force can change the system's configuration, so energy may go into internal or potential energy instead of only kinetic energy.

For friction, the mechanical energy lost is often modeled as:

$\Delta E_{\mathrm{mech}} = F_f d \cos\theta$

When friction acts opposite the motion, $\theta = 180^\circ$, so $\Delta E_{\mathrm{mech}} = -F_f d$, meaning mechanical energy drops by $F_f d$. That energy usually becomes thermal energy and sound.

Work from a Force-Displacement Graph

Work equals the area under a graph of the parallel force component $F_{\parallel}$ versus displacement $d$.

For a constant parallel force:

• The area is a rectangle with height $F_{\parallel}$ and width $d$.
• Work $= F_{\parallel} \times d$.

For a variable force:

• Find the area under the curve, often by breaking it into rectangles and triangles.
• Areas above the axis are positive work; areas below the axis are negative work.

How to Use This on the AP Physics 1 Exam

Problem Solving

• Identify the force, the displacement, and the angle between them before plugging into $W = Fd\cos\theta$.
• Check the sign: ask whether the parallel force component helps or opposes the motion. Negative work means energy leaves the system.
• For net work, either add the work from each force or use the work-energy theorem and solve for the unknown.
• When a force varies, switch to the area-under-the-graph method instead of a single $Fd\cos\theta$ calculation.

Free Response

• State the principle you are using (work-energy theorem, definition of work) before the algebra so your reasoning is clear.
• When a graph is involved, show how you split the area into shapes and include negative regions.
• For explanation parts, connect signs and components to the physics: say why a perpendicular force does no work or why friction removes mechanical energy.

Common Trap

• Do not multiply the full force magnitude by displacement when there is an angle; use only the parallel component.
• A force can be large and still do zero work if it is perpendicular to the motion or if nothing moves.

Common Misconceptions

• "Any applied force does work." Only the component parallel to the displacement does work, and if there is no displacement, the work is zero no matter how hard you push.
• "Work is a vector." Work is a scalar. It has a sign (positive, negative, or zero), but no direction.
• "Negative work means the object speeds up backward." Negative work just means energy is being removed from the system, which slows the object or stores energy elsewhere.
• "A single object can store potential energy by itself." Potential energy belongs to a system of interacting objects through a conservative force, not to one lone object.
• "Friction always equals the change in total energy." Friction reduces mechanical energy by $F_f d$, but that energy is not harmed; it usually becomes thermal energy and sound.
• "Normal force always does work because it is large." A normal force perpendicular to the motion does zero work, regardless of its size.

Practice Problem 1: Work from a Force-Displacement Graph

Solution

The work is the area under the $F_{\parallel}$ vs. $x$ graph.

From $0$ m to $2$ m: $W_1 = (4\,\text{N})(2\,\text{m}) = 8\,\text{J}$

From $2$ m to $5$ m: $W_2 = (0\,\text{N})(3\,\text{m}) = 0\,\text{J}$

From $5$ m to $7$ m: $W_3 = (-2\,\text{N})(2\,\text{m}) = -4\,\text{J}$

So the total work is: $W_{total} = 8\,\text{J} + 0\,\text{J} - 4\,\text{J} = 4\,\text{J}$

Practice Problem 2: Work-Energy Theorem Application

Solution

Use the work-energy theorem: $W_{net} = \Delta K$

The change in kinetic energy is: $\Delta K = \frac{1}{2}(2000\,\text{kg})(0\,\text{m/s})^2 - \frac{1}{2}(2000\,\text{kg})(15\,\text{m/s})^2$

$\Delta K = 0 - (1000)(225)\,\text{J} = -225{,}000\,\text{J}$

The work done by the braking force is: $W = F \times d \times \cos(180^\circ) = -F \times d$

Setting this equal to the change in kinetic energy: $-F \times 30\,\text{m} = -225{,}000\,\text{J}$ $F = \frac{225{,}000\,\text{J}}{30\,\text{m}} = 7{,}500\,\text{N}$

The average braking force has magnitude 7,500 N and points opposite the car's motion.

Practice Problem 3: Work Done by Multiple Forces

Solution

Applied force work: $W_{\text{applied}} = Fd\cos\theta = (30\,\text{N})(10\,\text{m})\cos 25^\circ \approx 272\,\text{J}$

Friction work: $W_f = Fd\cos 180^\circ = (5.4\,\text{N})(10\,\text{m})(-1) = -54\,\text{J}$

Normal force and weight do zero work because they are perpendicular to the horizontal displacement.

So the net work is: $W_{\text{net}} = 272\,\text{J} - 54\,\text{J} = 218\,\text{J}$

Related AP Physics 1 Guides

• 3.3 Potential Energy
• 3.4 Conservation of Energy
• 3.5 Power
• 3.1 Translational Kinetic Energy