Work, Energy, and Power: AP Physics 1 Unit 3

Why this unit matters

Energy methods often solve complex motion problems more directly than force-by-force kinematics.

What you will learn

Compute work done by constant forces using force, displacement, and angle.
Apply conservation of mechanical energy when assumptions are valid.
Interpret power as the rate of energy transfer in physical and engineering contexts.

Understand the core ideas

Work and energy provide a state-based approach to motion problems. For a constant force, work is W = Fd cos(theta), where theta is the angle between force and displacement. Positive work adds energy to the system, negative work removes it, and zero work occurs for perpendicular force-displacement geometry. Kinetic energy is $K = 0.5mv^2$ , and gravitational potential energy near Earth is often modeled as U_g = mgh relative to a chosen reference level.

Mechanical energy conservation applies when only conservative interactions exchange energy within the chosen system. In that case, K + U stays constant between states. If friction, drag, or external pushes are relevant, include nonconservative work terms rather than forcing a conservation equation that does not fit assumptions. Power quantifies transfer rate with P = W/delta t or P = Fv for constant force parallel to velocity.

The most important decision is system boundary selection. If Earth and object are both inside the system, gravitational transfer is internal and represented by potential energy change. If not, gravity is external work. Writing this choice before equations avoids double counting and improves scoring clarity.

Key terms

work: Energy transfer by force through displacement, W = Fd cos(theta) for constant force.
kinetic energy: Energy associated with motion $, K = 0.5mv^2$ .
mechanical energy: Sum of kinetic and potential energies for a defined system.
power: Rate of energy transfer, commonly measured in watts.

Speed at the bottom of a frictionless hill

A 2.0 kg cart starts from rest 5.0 m above the bottom of a frictionless track. Let bottom height be zero.

Write conservation of mechanical energy: K1 + U1 = K2 + U2.
Substitute known values: 0 + mgh $= 0.5 m v^2 + 0$ .
Insert numbers: $(2.0)(9.8)(5.0) = 0.5(2.0)v^2$ , so $98 = v^2$ .
Solve and apply sign context: v = $\sqrt{98}$ = 9.9 m/s, taking the positive speed magnitude.
Check units before solving: mgh and $0.5mv^2$ are both joules, confirming dimensional consistency.

Result: The cart reaches the bottom at about 9.9 m/s.

A common misconception

Claim: Heavier objects always reach higher speed than lighter ones from the same frictionless drop.

Correction: In the ideal model, mass cancels in mgh $= 0.5mv^2$ , so final speed depends on height change, not mass.

Lessons in this unit

Work by constant forcesDetermine positive, negative, or zero work from force-displacement geometry.
Kinetic and potential energyRelate speed and position changes to energy changes in a system.
Conservation of energyUse system boundaries to include or exclude nonconservative work.
Power and efficiencyAnalyze how quickly energy is transferred or transformed.

Study task

Model a roller coaster segment with given heights and one friction region. Compute speeds at key points and compare energy before and after the friction segment.

Unit checkpoint

How much work is done by a 10 N force parallel to a 3.0 m displacement?

W = Fd cos(theta). Here theta = 0 deg, so W = (10)(3.0)(1) = 30 J.

Work, Energy, and Power