Unit 3 Notes: Energy Methods and Power in AP Physics 1

Conservation of Energy

What “energy” means in mechanics

Energy is a property of a system that measures its ability to cause change—most often to cause motion or to deform something. In AP Physics 1 mechanics, you usually track energy because it lets you predict speeds, heights, and spring compressions without needing to analyze every moment of acceleration.

A crucial idea: energy is not “stored in an object” in a simple way; it is stored in configurations and interactions. For example, gravitational potential energy belongs to the Earth–object system, not to the object alone.

In this unit, the most common energy forms are:

  • Kinetic energy (energy of motion):

K = \frac{1}{2}mv^2

Here m is mass and v is speed.

  • Gravitational potential energy near Earth’s surface (energy from height):

U_g = mgh

Here g is gravitational field strength (near Earth, approximately constant), and h is vertical height measured from a chosen zero level.

  • Spring (elastic) potential energy:

U_s = \frac{1}{2}kx^2

Here k is the spring constant and x is the stretch/compression from the spring’s relaxed length.

A common misconception is to treat these as unrelated “equations to plug into.” Instead, treat them as different bookkeeping ways to track how interactions (gravity, springs) and motion trade energy.

The core principle: conservation of energy (and when it applies)

Conservation of energy means the total energy of an isolated system cannot change—energy can be transferred or transformed, but it cannot be created or destroyed.

In AP Physics 1 problems, you often focus on mechanical energy:

  • Mechanical energy is the sum of kinetic energy and potential energies you’re tracking:

E_{mech} = K + U_g + U_s

Mechanical energy is conserved when only conservative forces do work within the system. Conservative forces (like gravity and ideal spring forces) have a special property: the work they do depends only on initial and final positions, not on the path.

When nonconservative forces (like kinetic friction, air resistance, or an external push/pull) are involved, mechanical energy is generally not conserved. But total energy is still conserved—you just have to account for where the “missing” mechanical energy goes (usually into thermal energy of surfaces and the environment).

Systems thinking: the most important setup step

Energy methods depend on choosing a system. Your system choice determines what counts as “internal” energy changes versus “external” work.

  • If a force comes from inside the system (an interaction between parts of the system), it can convert energy between forms within your energy accounting.
  • If a force comes from outside the system, it typically transfers energy into or out of the system via work.

This is why, for gravity problems, including Earth in the system is so helpful: gravity becomes an internal interaction and you can use U_g.

Work and the energy equation you’ll actually use

Work is energy transferred by a force acting through a displacement. In energy conservation problems, the key relationship is:

  • Work–energy theorem (net work changes kinetic energy):

W_{net} = \Delta K

  • General energy accounting form (very useful for AP Physics 1):

K_i + U_i + W_{ext} = K_f + U_f

Here:

  • K_i, K_f are initial and final kinetic energies
  • U_i, U_f are initial and final potential energies (gravity and/or spring)
  • W_{ext} is work done on the system by forces external to the system (often friction, a push, or tension from an external agent)

If you choose a system so that all forces are conservative and internal, then W_{ext}=0 and you get conservation of mechanical energy:

K_i + U_i = K_f + U_f

How to think about signs (where students often go wrong)

A common source of error is sign confusion. A reliable approach is:

  • Compute energies from definitions (they are nonnegative for K and U_s).
  • Let U_g = mgh with a clearly defined zero height; then U_g can be positive, zero, or negative depending on your chosen reference.
  • Compute work by friction or applied forces carefully: kinetic friction often removes mechanical energy from the tracked forms, so its work is often negative if it opposes motion.

Conservative forces and potential energy

A force is conservative if you can define a potential energy function for it. In AP Physics 1, the two big conservative forces are gravity (near Earth) and springs.

Why this matters: if only conservative forces do work (and no external work is done), you can solve problems by comparing two states—no need to find acceleration as a function of time.

Nonconservative forces: friction and “lost” mechanical energy

Kinetic friction converts mechanical energy into thermal energy of the surfaces. In many AP Physics 1 problems, you don’t calculate temperature changes; you just account for energy leaving mechanical forms.

If you treat friction as external work on your chosen system, you include W_{ext} (typically negative). A common model for kinetic friction magnitude is:

f_k = \mu_k N

If friction is approximately constant and opposite the motion over distance d, then the work done by friction is:

W_f = -f_k d

(negative because friction opposes the displacement).

A frequent mistake is to assume friction “uses up energy so energy isn’t conserved.” Total energy is always conserved; friction is an energy transfer/transform mechanism.

Representations you should be fluent with: energy bar charts and diagrams

AP questions often present (or expect you to use) qualitative representations:

  • Energy bar charts: compare K, U_g, U_s (and sometimes thermal) at two moments.
  • System schema/interaction diagrams: identify which interactions are internal (becoming potential energy) versus external (becoming work).

A good habit: before writing equations, describe the story in energy terms—“gravitational potential decreases, kinetic increases, some energy becomes thermal due to friction.” This prevents blindly plugging.

Worked example 1: frictionless ramp (mechanical energy conserved)

A block of mass m starts from rest at height h and slides down a frictionless ramp. Find its speed at the bottom.

Conceptual setup: With no friction and ignoring air resistance, mechanical energy is conserved. Choose the system as block + Earth so gravity is internal and becomes U_g.

At the top:

  • K_i = 0
  • U_{g,i} = mgh

At the bottom (take h=0 there):

  • U_{g,f} = 0
  • K_f = \frac{1}{2}mv^2

Use conservation of mechanical energy:

K_i + U_i = K_f + U_f

0 + mgh = \frac{1}{2}mv^2 + 0

Solve:

v = \sqrt{2gh}

What to notice: mass cancels—so for the same drop height, all masses reach the same speed (in this ideal model). Students often find that surprising because they mix up force size with final speed; energy depends on height change, not on time taken.

Worked example 2: ramp with kinetic friction (mechanical energy not conserved)

A block of mass m starts from rest at height h and slides down a ramp of length L with kinetic friction coefficient \mu_k. Find its speed at the bottom.

Conceptual setup: Gravity increases kinetic energy, but friction removes mechanical energy. Choose system as block + Earth (so gravity is handled by U_g). Treat friction as external work on the system.

Energy equation:

K_i + U_i + W_f = K_f + U_f

Compute each term:

  • K_i = 0
  • U_i = mgh
  • U_f = 0
  • W_f = -f_k L = -\mu_k N L

If the ramp is at angle \theta, then N = mg\cos\theta (because normal balances the perpendicular component of weight), so:

W_f = -\mu_k mg\cos\theta L

Plug in:

mgh - \mu_k mg\cos\theta L = \frac{1}{2}mv^2

Cancel m and solve:

v = \sqrt{2g\left(h - \mu_k \cos\theta\, L\right)}

Common pitfall: using h as the ramp length or using L as the vertical drop. Energy from gravity depends on vertical height change; friction work depends on the distance traveled along the surface.

Worked example 3: spring launcher (gravity + spring + kinetic)

A spring with constant k is compressed by distance x and launches a block of mass m on a horizontal, frictionless surface. Find the block’s speed when the spring returns to its relaxed length.

Conceptual setup: Spring potential energy converts into kinetic energy. With no friction, mechanical energy is conserved.

Initial:

  • K_i = 0
  • U_{s,i} = \frac{1}{2}kx^2

Final (spring relaxed):

  • U_{s,f} = 0
  • K_f = \frac{1}{2}mv^2

Conservation:

\frac{1}{2}kx^2 = \frac{1}{2}mv^2

v = x\sqrt{\frac{k}{m}}

Common pitfall: confusing x (compression from equilibrium) with the total distance the block travels after leaving the spring.

Exam Focus

  • Typical question patterns:
    • “Object moves from A to B” problems where you must choose a system and decide whether K + U is conserved or whether you must include external work (often friction).
    • Mixed-energy problems involving springs and gravity (block launched by spring up a ramp; find max height or compression).
    • Qualitative bar-chart questions: compare energies at two points, or identify which interactions cause energy transfers.
  • Common mistakes:
    • Treating friction as making energy “not conserved” instead of accounting for it as external work or thermal energy.
    • Using ramp length as height in mgh, or forgetting that only vertical height change affects gravitational potential energy.
    • Choosing inconsistent reference heights for h across the problem (changing your zero level mid-solution).

Power

What power is (and why you care)

Power is the rate at which energy is transferred or transformed. Energy tells you “how much,” but power tells you “how fast.” Two machines might do the same amount of work (same energy transfer) but at different rates—one feels “more powerful” because it delivers that energy faster.

In everyday terms, a light bulb’s wattage is power: it tells you how quickly electrical energy is being converted into light and thermal energy.

The core definitions

The most fundamental definition is average power:

P_{avg} = \frac{W}{\Delta t}

Here:

  • P_{avg} is average power
  • W is work done (energy transferred)
  • \Delta t is the time interval

The SI unit is the watt (W):

1\ \text{watt} = 1\ \text{joule per second}

So if you do 300\ \text{J} of work in 2\ \text{s}, your average power is 150\ \text{W}.

Instantaneous power and the connection to force and speed

Sometimes AP Physics 1 asks about power at an instant rather than over a time interval. If a force acts on an object that is moving, instantaneous power (in 1D, or when force is parallel to velocity) can be written as:

P = Fv

Here F is the component of force in the direction of motion and v is the instantaneous speed.

Why this makes sense: work over a small displacement is approximately W \approx F\Delta x, and dividing by time gives \frac{W}{\Delta t} \approx F\frac{\Delta x}{\Delta t} = Fv.

Common misconception: students often use P = Fv even when the applied force is not parallel to the velocity. The more general relationship uses the component of force along the motion; if the force is perpendicular to the motion, it does no work and contributes no power.

Lifting at constant speed: a classic power situation

When you lift an object of mass m straight up at constant speed, the applied force is approximately its weight mg (ignoring air resistance and assuming no acceleration). The work to lift through height h is mgh, so average power is:

P_{avg} = \frac{mgh}{\Delta t}

If the object is lifted at constant speed v, then h = v\Delta t, which leads to:

P = mgv

This is a powerful connection: for a given weight, lifting faster requires more power.

Power in energy conservation problems

Power often appears as an extra layer on energy conservation:

  • Energy equations tell you the change in energy needed.
  • Power tells you how quickly that energy change happens.

A typical two-step approach:

  1. Use energy to find required work/energy transfer (for example, increase in gravitational potential energy).
  2. Use P_{avg} = \frac{W}{\Delta t} to solve for time or power.

Efficiency (when not all input power becomes useful output)

Real devices waste some input energy (often as thermal energy and sound). Efficiency measures the fraction of input converted into useful output:

\text{efficiency} = \frac{P_{out}}{P_{in}}

(If asked for percent efficiency, multiply by 100%.)

A common mistake is mixing energy and power in the ratio; you must compare power-to-power or energy-to-energy over the same interval.

Worked example 1: power to climb stairs

A student of mass m = 60\ \text{kg} climbs a vertical height h = 4.0\ \text{m} in \Delta t = 5.0\ \text{s}. Find the student’s average power output against gravity.

Conceptual setup: The useful work is the increase in gravitational potential energy.

Work done against gravity:

W = mgh

Average power:

P_{avg} = \frac{mgh}{\Delta t}

Substitute:

P_{avg} = \frac{(60)(9.8)(4.0)}{5.0}

Compute:

P_{avg} \approx 470\ \text{W}

Interpretation: This is the rate at which chemical energy is being converted into gravitational potential energy (plus additional wasted energy not included in this idealized calculation).

Worked example 2: constant-speed towing and P = Fv

A car pulls a trailer at constant speed v = 20\ \text{m/s}. The resistive force opposing motion is F = 500\ \text{N}. What power must the car deliver just to overcome this resistance (ignoring other losses)?

Conceptual setup: At constant speed, the car’s forward force matches the resistive force. The power needed to supply energy at the rate it is dissipated is P = Fv.

P = Fv = (500)(20) = 1.0\times 10^4\ \text{W}

So:

P = 10000\ \text{W}

Common pitfall: using P = \frac{W}{\Delta t} but not knowing how to get work. Here, the force acts continuously; P = Fv is the direct route.

Worked example 3: using power to find time with an energy change

An elevator lifts a total mass m = 800\ \text{kg} a height h = 12\ \text{m}. If the motor delivers constant power P = 1.6\times 10^4\ \text{W} to the elevator (idealized), how long does the lift take?

Conceptual setup: The required energy increase is gravitational potential energy. Time is energy divided by power.

Required work:

W = mgh

Time from P = \frac{W}{\Delta t}:

\Delta t = \frac{W}{P} = \frac{mgh}{P}

Substitute:

\Delta t = \frac{(800)(9.8)(12)}{1.6\times 10^4}

Compute:

\Delta t \approx 5.9\ \text{s}

What to watch: This ignores the elevator’s changing kinetic energy (acceleration phases). On AP problems, if they say “constant speed,” you can usually ignore kinetic energy changes; if they mention starting/stopping, you may need to account for additional energy temporarily stored as kinetic.

How power shows up in graphs and reasoning

You may see questions that connect power to slopes:

  • On an energy vs. time graph, the slope represents power because P = \frac{\Delta E}{\Delta t}.
  • On a work vs. time graph, the slope is also power.

Even if a problem doesn’t explicitly say “graph,” this interpretation can help you sanity-check: if energy is increasing quickly, power must be large.

Exam Focus

  • Typical question patterns:
    • “How long does it take” questions where you compute an energy change (often mgh) and divide by a given power.
    • Constant-speed motion with a known resistive force, asking for required power using P = Fv.
    • Conceptual comparisons: same work done in different times, or same power for different masses/heights.
  • Common mistakes:
    • Confusing energy with power: stating something like “the power is mgh” (that’s energy, not rate).
    • Using P = Fv when force is not along the direction of motion (only the parallel component contributes).
    • Ignoring “useful output” vs. “input” in efficiency problems and mixing units (watts vs. joules).