AP Physics 1  ·  Unit 3: Work, Energy & Power  ·  Lesson 3.2

Deep Dive: Work

🔬 Deep Dive
This is your textbook for this topic. Take your time. Read it more than once.
3.2.A.1Concept

Work as Energy Transfer

Work is not effort. In physics, work has a precise definition: it is the amount of energy transferred into or out of a system by a force acting over a displacement. You can push against a wall all day and do zero work — because the wall doesn't move.

Three things are required for work to be done: a force, a displacement, and the force must have a component along the displacement. All three conditions must be met. Missing any one of them means zero work.

🔑Work is the mechanism of energy transfer. It's the only way energy moves into or out of a system in mechanics. If you want to change an object's kinetic energy, something must do work on it.

Work is measured in Joules (J) — the same unit as energy, because work is energy transfer. Work can be positive (energy added to the system), negative (energy removed), or zero (no transfer).

3.2.A.2ConceptMath

W = F‖ · d

The work done by a force is the product of the component of force parallel to the displacement and the magnitude of the displacement:

W = F‖ · d

When a force is applied at an angle θ to the direction of motion, the parallel component is F cos θ. So the formula becomes:

W = Fd cos θ

Where θ is the angle between the force vector and the displacement vector. This is the form you'll see on the AP equation sheet — but understanding it as "only the parallel component" is more useful than memorizing the cosine.

Set the force, displacement, and angle. Watch the parallel component (the only part that does work) change as the angle increases toward 90°.

Force (F)50N
Displacement (d)8m
Angle (θ)30°
d = 8 mF = 50NF‖ = 43.3NF⊥=25.0N30°
F‖ = F cos θ
43.3 N
does work
F⊥ = F sin θ
25.0 N
does NO work
W = F‖ × d
346.4 J
energy transferred

Drag angle to 90° — work drops to zero. The force is still there, but none of it is aligned with motion. No energy transferred.

ExampleWorked Example — Pushing at an Angle

A person pushes a 20 kg box across the floor with a force of 80 N at 35° below horizontal. The box moves 6 m horizontally. How much work does the person do?

3.2.A.3Concept⚠ Watch Out

Perpendicular Forces Do Zero Work

When a force acts exactly perpendicular to the displacement — θ = 90° — cos 90° = 0, so W = 0. The force transfers no energy. This is one of the most important results in the unit.

Normal force on horizontal surface
Acts upward; motion is horizontal. 90° angle. Zero work.
Centripetal force in circular motion
Always points inward (toward center); velocity is tangential. 90° angle. Zero work.
Tension on pendulum bob
Along the string (toward pivot); velocity is perpendicular to string. 90° angle. Zero work.
Magnetic force on moving charge
Always perpendicular to velocity by definition. Zero work — ever.
⚠️Centripetal force never does work. An object in uniform circular motion has constant speed — its kinetic energy never changes — because the centripetal force is always perpendicular to velocity. This is why planets can orbit indefinitely without any external energy input.
3.2.A.4MathConcept

The Work-Energy Theorem

The work-energy theorem states that the net work done on an object equals its change in kinetic energy:

W_net = ΔK = Kf − K₀

This is one of the most powerful results in classical mechanics. It connects what forces do (work) directly to what motion becomes (kinetic energy), bypassing the need to track force and acceleration at every instant.

🔑The work-energy theorem uses net work — the total work done by all forces combined. Individual forces may do positive or negative work; what matters for changing KE is their sum.

Set an object's initial state and the net work done on it. Watch the work-energy theorem connect work directly to the change in kinetic energy.

Mass (m)5 kg
Initial speed v₀4 m/s
Net work W_net+60 J
K₀ (initial)40.0 J
Kf (final)100.0 J
W_net = +60 J
ΔK = Kf − K₀ = 100.040.0 = 60.0 J
vf = √(2Kf/m) = 6.32 m/s

Try setting W_net negative — it removes energy. At some point KE hits zero and the object stops. Kinetic energy can't go below zero.

ExampleGuided Example — Finding Final Speed via W-E Theorem

A 3 kg object starts at rest. A net force does 96 J of work on it. What is its final speed?

Step 1Apply the work-energy theorem
W_net = ΔK = Kf − K₀. Since it starts at rest, K₀ = 0. So Kf = W_net = 96 J.
3.2.A.5Concept

Conservative vs. Nonconservative Forces

Not all forces behave the same way when an object moves along a path. This distinction matters enormously for how we account for energy.

Conservative

The work done depends only on the starting and ending positions — not on the path taken between them.

Gravity, spring force, electric force
Associated with potential energy. If the object returns to its start, the net work done by a conservative force is zero.
Nonconservative

The work done depends on the path taken. A longer or rougher path means more energy lost.

Kinetic friction, air resistance, tension in a rope that generates heat
Dissipates energy as heat or sound. Cannot be stored as potential energy. Energy is permanently lost from the mechanical system.
💡This distinction becomes critical in Lesson 3.4. When only conservative forces do work, mechanical energy is conserved exactly. When nonconservative forces do work, some mechanical energy is lost to heat — and you need to account for that loss explicitly.
3.2.A.6Math

Graphical Area Analysis

When the force is constant, W = F‖ · d is straightforward. But when force varies with displacement — like a spring — you need a different approach. On a force vs. displacement graph, the total work done is the area under the curve.

Displacement (m)Force (N)RectangleW = F × dTriangleW = ½F_max·dconstant Fspring (F = kx)
🔑For a constant force: area = rectangle = F × d.
For a spring (F = kx, linear): area = triangle = ½ × F_max × d = ½kd². This is exactly the elastic potential energy formula you'll see in Lesson 3.3.
ExampleWorked Example — Work from a Graph

A force vs. displacement graph shows a straight line from (0, 0) to (4 m, 24 N). What is the total work done over this displacement?

← Back to Lesson 3.2Next: Lesson 3.3 →Potential Energy — storing energy for later.
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