Energy — AP Physics 1 Phys 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Work-energy theorem, kinetic and potential energy, conservation of mechanical energy, average and instantaneous power, and spring potential energy, with exam-focused worked examples and common mistake warnings.

You should already know: Algebra 2, basic trig, no calculus required.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 1 style for educational use. They are not reproductions of past College Board papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Energy?

Energy is a conserved scalar quantity that describes the capacity of a system to perform work on other systems, measured in joules (J, equivalent to $N\cdot m$). As a scalar, it has no direction, only magnitude, and can be converted between different forms but never created or destroyed, per the first law of thermodynamics. Per the 2024-25 AP Physics 1 CED, energy topics make up 12-18% of your total exam score, appearing in both multiple-choice and multi-part free-response questions, often paired with kinematics, forces, or circular motion.

2. Work-Energy Theorem

The work-energy theorem is the foundational link between force, motion, and energy, allowing you to solve kinematics problems without calculating acceleration or time intervals if you know energy changes. First, define work done by a constant force: work $W$ is the product of the component of force parallel to displacement, the magnitude of displacement, and the cosine of the angle between the force and displacement vectors: $$W=F\Delta x\cos \theta$$ Net work $W_{net}$ is the sum of work done by all individual forces acting on an object, or equivalently the work done by the net force on the object. The work-energy theorem states that net work equals the change in the object’s kinetic energy: $$W_{net}=\Delta KE=K{E}_{final}-K{E}_{initial}$$ This is derived directly from Newton’s second law $F_{net}=ma$ and the kinematic equation $v^2=u^2+2a\Delta x$, rearranged to eliminate acceleration and time.

Worked Example

A 2kg box is pushed across a frictionless floor with a constant 10N force parallel to its displacement of 5m. What is its final speed if it starts from rest?

1. Calculate net work: the force is parallel to displacement, so $\theta =0$, $\cos \theta =1$, $W_{net}=10N\times 5m=50J$
2. Set equal to change in kinetic energy: $50J=\frac{1}{2}(2kg)v_f^2-0$
3. Solve for $v_f$: $v_f^2=50$, so $v_f\approx 7.1m/s$ Exam tip: Examiners often test that only net work matters: if friction is present, subtract the work done by friction (which is negative, as it opposes motion) to get the correct net work value.

3. Kinetic and Potential Energy

Energy exists in two broad categories relevant to AP Physics 1: kinetic energy (energy of motion) and potential energy (stored energy due to position or configuration).

Kinetic Energy (KE)

Kinetic energy is the energy of a moving object, calculated as: $$KE=\frac{1}{2}m{v}^2$$ Where $m$ is mass in kg and $v$ is speed in m/s. KE is always non-negative, as speed is squared, so direction of motion does not affect its value: doubling speed quadruples KE, a common exam test point.

Gravitational Potential Energy ($U_g$)

Gravitational potential energy is stored energy due to an object’s vertical position relative to a chosen reference point, calculated as: $$U_g=mgh$$ Where $g=9.8m/s^2$, and $h$ is vertical height in meters relative to your chosen $h=0$ reference point. You can set $h=0$ anywhere, as long as you use the same reference for all steps of a problem: only changes in $U_g$ affect calculations, not absolute values.

Worked Example

A 3kg ball is held 2m above a table, which is 1m above the floor. Calculate $U_g$ relative to the table and relative to the floor.

1. Relative to the table ($h=0$ at table surface): $U_g=3kg\times 9.8m/s^2\times 2m=58.8J$
2. Relative to the floor ($h=0$ at floor): $U_g=3kg\times 9.8m/s^2\times 3m=88.2J$ Note: If you drop the ball, the change in $U_g$ when it hits the table is 58.8J regardless of reference point, so final speed calculations will be identical.

4. Conservation of Mechanical Energy

Mechanical energy $E_{mech}$ is the sum of kinetic and potential energy in a system: $E_{mech}=KE+U$ (where $U$ includes all forms of potential energy in the system). The law of conservation of mechanical energy states that if only conservative forces (gravity, spring force) do work on a system, with zero work done by non-conservative forces (friction, air resistance, applied external pushes/pulls), total mechanical energy remains constant: $$K{E}_i+U_i=K{E}_f+U_f$$ Conservative forces have path-independent work: the work done by gravity to move a box from a height of 5m to 0m is the same if it falls straight down or slides down a ramp, for example. Non-conservative forces have path-dependent work, and dissipate energy as heat or sound. If non-conservative forces do work, add a non-conservative work term to the equation: $$K{E}_i+U_i+W_{nonconservative}=K{E}_f+U_f$$ $W_{nonconservative}$ is negative for forces like friction that remove energy from the system.

Worked Example

A 5kg pendulum is released from rest at a height of 0.8m above its lowest point. What is its speed at the lowest point, ignoring air resistance?

1. Choose $h=0$ at the lowest point, so $U_{g,f}=0$, and initial $K{E}_i=0$ (released from rest)
2. Initial energy: $U_{g,i}=5kg\times 9.8m/s^2\times 0.8m=39.2J$
3. Final energy: $K{E}_f=\frac{1}{2}(5kg)v_f^2$
4. Equate initial and final energy: $39.2J=2.5v_f^2$, so $v_f\approx 4.0m/s$ Exam tip: Examiners frequently add friction to this problem type to test if you remember to include the negative non-conservative work term, reducing the final speed.

5. Power — Average and Instantaneous

Power is the rate at which work is done, or the rate at which energy is transferred between systems, measured in watts (W, equivalent to J/s).

Average Power

Average power is the total work done over a finite time interval, calculated as: $$P_{avg}=\frac{\Delta W}{\Delta t}$$ For constant force, this can also be written as $P_{avg}=F{v}_{avg}\cos \theta$, where $v_{avg}$ is average speed over the interval.

Instantaneous Power

Instantaneous power is the power delivered at a specific moment in time, calculated as: $$P_{inst}=F{v}_{inst}\cos \theta$$ Where $v_{inst}$ is instantaneous speed, and $\theta$ is the angle between the force and velocity vectors at that moment.

Worked Example

A 1000kg car accelerates from rest to 20m/s in 10s on a flat road, ignoring friction. Calculate average power output of the engine, and instantaneous power at $t=10s$.

1. Work done by the engine equals the change in KE: $\Delta KE=\frac{1}{2}(1000kg)(20m/s{)}^2=200,000J$
2. Average power: $P_{avg}=\frac{200,000J}{10s}=20,000W=20kW$
3. Constant acceleration: $a=\frac{20m/s}{10s}=2m/s^2$, so engine force $F=ma=2000N$
4. Instantaneous power at $t=10s$: $P_{inst}=2000N\times 20m/s=40,000W=40kW$ Exam trap: A common mistake is using average power when the question asks for power at a specific speed, leading to half the correct value for constant acceleration problems.

6. Spring Potential Energy

Spring potential energy (elastic potential energy, $U_s$) is the energy stored in a stretched or compressed spring. First recall Hooke’s Law, which describes the restoring force of a spring displaced $x$ from its equilibrium (unstretched) position: $$F_s=-kx$$ Where $k$ is the spring constant (units N/m, describing spring stiffness), and the negative sign indicates the force acts opposite to displacement to return the spring to equilibrium. Spring potential energy is derived from the work done to stretch or compress the spring, and is calculated as: $$U_s=\frac{1}{2}k{x}^2$$ Note that $x$ is squared, so energy is identical for equal amounts of stretch or compression from equilibrium, and $x$ is displacement from equilibrium, not total spring length.

Worked Example

A spring with $k=400N/m$ is compressed by 0.2m from equilibrium, then released to push a 0.5kg block across a frictionless surface. What is the speed of the block when it leaves the spring?

1. Conservation of mechanical energy: initial energy is all spring PE, final energy is all KE of the block
2. Initial $U_s=\frac{1}{2}(400N/m)(0.2m{)}^2=8J$
3. Final $KE=\frac{1}{2}(0.5kg)v_f^2=0.25v_f^2$
4. Equate: $8J=0.25v_f^2$, so $v_f\approx 5.7m/s$ Exam note: Questions often combine springs with gravity, e.g. a mass hanging on a spring: remember to include both gravitational PE and spring PE in your energy balance for these cases.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Using the angle between force and the horizontal instead of force and displacement in work calculations. Why students do it: They carry over angle conventions from force component questions. Correct move: Always measure $\theta$ between the force vector and displacement vector; if force opposes motion, $\theta =180^{\circ }$, $\cos \theta =-1$, so work is negative.
• Wrong move: Setting inconsistent $h=0$ reference points for initial and final states in potential energy calculations. Why students do it: They default to ground level as $h=0$ even if the system never reaches the ground. Correct move: Choose $h=0$ at the lowest point of the system’s motion to simplify calculations, and use that reference for all states in the problem.
• Wrong move: Assuming mechanical energy is always conserved, even when friction or applied external forces are present. Why students do it: They overgeneralize the conservation rule after practicing frictionless problems. Correct move: First list all forces doing work: if any non-conservative force acts, add the $W_{nonconservative}$ term to your energy balance.
• Wrong move: Using total spring length instead of displacement from equilibrium in spring potential energy calculations. Why students do it: They confuse stretch/compression with total spring length. Correct move: Subtract the unstretched equilibrium length from the current spring length to get $x$, and always square $x$ so its sign does not matter.
• Wrong move: Forgetting that spring potential energy is always positive, even when the spring is compressed. Why students do it: They carry over the negative sign from Hooke’s Law force calculations. Correct move: The square of displacement in $U_s$ eliminates the sign, so energy is identical for equal stretch and compression.

8. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

A 4kg box slides down a 30° incline of length 10m. The coefficient of kinetic friction between the box and incline is 0.2. If the box starts from rest at the top of the incline, what is its speed at the bottom? A) 7.3 m/s B) 8.1 m/s C) 9.2 m/s D) 10.0 m/s

Solution

1. Calculate non-conservative work from friction: height of incline $h=10\sin 30^{\circ }=5m$, normal force $N=mg\cos 30^{\circ }\approx 33.9N$, friction force $f=\mu N\approx 6.78N$, work done by friction $W_f=f\Delta x\cos 180^{\circ }=6.78\ast 10\ast (-1)=-67.8J$
2. Energy balance: $U_{g,i}+W_f=K{E}_f$ (set $h=0$ at bottom, $K{E}_i=0$, $U_{g,f}=0$)
3. Plug in values: $4\ast 9.8\ast 5-67.8=\frac{1}{2}\ast 4\ast v^2$ → $196-67.8=2v^2$ → $v^2=64.1$ → $v\approx 8.0m/s$, closest to option B.

Question 2 (Free Response Part A)

A student uses a spring launcher to fire a 0.1kg ball vertically upward. The spring has $k=500N/m$ and is compressed 0.1m before launch. Air resistance is negligible. Calculate the maximum height the ball reaches above the launch position (where it leaves the spring).

Solution

1. Energy balance: initial spring PE converts entirely to gravitational PE at maximum height (KE=0 at max height)
2. $\frac{1}{2}k{x}^2=mgh$
3. Solve for $h$: $h=\frac{k{x}^2}{2mg}=\frac{500\ast (0.1{)}^2}{2\ast 0.1\ast 9.8}=\frac{5}{1.96}\approx 2.6m$

Question 3 (Free Response Part B)

If the launcher’s motor does 15J of work over 0.5s to compress the spring, what is the average power output of the motor?

Solution

Average power = total work over time interval: $P_{avg}=\frac{\Delta W}{\Delta t}=\frac{15J}{0.5s}=30W$

9. Quick Reference Cheatsheet

10. What's Next

The energy concepts covered in this guide are foundational for 60% of the remaining AP Physics 1 syllabus, including rotational energy (which extends the work-energy theorem to rotating objects), momentum and collisions (where energy conservation distinguishes elastic and inelastic collisions), and simple harmonic motion (which relies on continuous conversion between kinetic, gravitational, and spring potential energy). You will also see energy combined with kinematics and forces in multi-part FRQs that make up 30% of your total exam score, so mastering these rules is non-negotiable for earning a 5 on AP Physics 1.

If you struggle with any of the worked examples, common pitfalls, or want to practice more exam-style questions tailored to your weak points, you can ask Ollie, our AI tutor, for personalized help at any time. You can also find more topic guides, full practice exams, and grading rubrics aligned to the 2024-25 AP Physics 1 CED on the OwlsPrep homepage.