AP Physics 1 / Topic

Work, Energy, and Power

Unit 3 reframes mechanics in terms of energy: work done by forces, kinetic and potential energy, the work-energy theorem, conservation of energy with and without friction, and power.

44 practice questions8 FRQs7 skill areas

What the exam asks

  • Compute work done by constant forces, including negative work and zero-work cases
  • Apply the work-energy theorem to connect net work and speed change
  • Use energy conservation across positions (ramps, pendulums, springs, projectiles)
  • Account for friction as mechanical energy converted to thermal energy
  • Calculate power as the rate of energy transfer

Key formulas and rules

Work

W = F·d·cosθ; area under F-x graph

Work-energy theorem

W(net) = ΔKE

Energy forms

KE = ½mv²; PE(grav) = mgh; PE(spring) = ½kx²

Conservation with friction

KE1 + PE1 = KE2 + PE2 + f·d

Power

P = W/t = F·v

Question bank breakdown

Multiple choice

44

Free response

8

Difficulty mix

12 easy · 24 medium · 8 hard

Skills covered

WorkConservation of EnergyWork Done by a ForceWork-Energy TheoremPowerPotential Energy (Gravitational and Elastic)Translational Kinetic Energy

Every question in the bank comes with a worked explanation, and most add a per-choice breakdown so you learn why each wrong answer is wrong - not just the key.

Sample question

From the free tier of the ScoreMint AP Physics 1 bank - try it, then check the answer.

A 4.0 kg4.0 \text{ kg} block slides across a rough horizontal surface. Its speed decreases from 6.0 m/s6.0 \text{ m/s} to 2.0 m/s2.0 \text{ m/s} over a distance of 4.0 m4.0 \text{ m}. What is the magnitude of the friction force acting on the block?

  • A. 8.0 N8.0 \text{ N}
  • B. 16 N16 \text{ N}
  • C. 32 N32 \text{ N}
  • D. 4.0 N4.0 \text{ N}
Show answer and explanation

Answer: B. The work-energy theorem states that the net work done on an object equals its change in kinetic energy: Wnet=ΔKEW_{net} = \Delta KE. Here, ΔKE=12(4.0)(2.02)12(4.0)(6.02)=872=64\Delta KE = \frac{1}{2}(4.0)(2.0^2) - \frac{1}{2}(4.0)(6.0^2) = 8 - 72 = -64 J. The only horizontal force is friction, and its work is Wf=fdW_f = -fd (negative because friction opposes displacement). Setting f(4.0)=64-f(4.0) = -64 J gives f=16f = 16 N. The negative sign of ΔKE\Delta KE confirms energy was removed from the block by friction.

Study tip

Choose energy methods when the question asks about speeds at two positions and forces are messy in between; choose Newton's laws when it asks about forces or acceleration at one instant.