Overview

How to Solve Any AP Physics Energy Question

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Jason Kuma

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This framework will show you a fool proof method to solve any energy-related question you encounter in physics.

Know Your Energies

AP Physics covers only 3 types of energy:

  1. Gravitational Potential Energy [katex] PE_g = mgh [/katex].
    • GPE depends solely on the height. If an object gains height, it gains GPE. If its looses height it looses GPE
  2. Kinetic Energy – [katex] KE = \frac{1}{2}mv^2 [/katex]
    • KE depends solely on the speed an object gains or looses.
    • Rotational kinetic energy follows a similar formula and applies to objects that rotate [katex] KE = \frac{1}{2}I\omega^2 [/katex]
  3. Spring Potential Energy – [katex] KE = \frac{1}{2}kx^2 [/katex]

Lastly every type of energy thats not the 3 types above is classified as Work Energy where [katex] W = Fd [/katex]. A common example is the work done by friction.

If you need a quick review of more energy concepts check this out energy speed review.

Just 3 steps

  1. Identify energies and visualize the system: either in your head or draw a diagram.
  2. Mark two points in the system: an initial starting point (point A) and an ending point (point B).
  3. Apply conservation of energy: [katex] E_A = E_B [/katex], which simply states the sum of all energy types at point A is equal to the sum of all of energy types at point B.

Apply it

Lets apply the energy framework to the problem below.

Open in UBQ
Question 1
Advanced
Mathematical
A \(2 \, \text{kg}\) model rocket is launched with a thrust force of \(275 \, \text{N}\) and reaches a height of \(90 \, \text{m}\), at which point the thrust cuts out, but the rocket continues moving at \(150 \, \text{m/s}\). What is the average air resistance force acting on the rocket during its ascent?

Step 1: Image a rocket launch. After some time the rocket will reach a height with a certain speed. Identify energies associated in this scenario. For example:

  1. The moving rocket implies kinetic energy.
  2. The rocket being at a certain height also implies potential energy.
  3. This energy must be coming from the rocket, thus there must be work done by the rocket.
  4. Lastly the rocket is pushing against air. This implies some work done by air resistance.

Step 2: The starting point A would be when the rocket begins to take off. At this point all the energy is stored in the rocket. The final point B would be when the rocket reaches a certain height. At this point the energy from the rocket transforms into the kinetic energy, potential energy, and work done by air resistance.

Step 3: Apply the conservation of energy:

[katex=display] E_A = E_B [/katex]

[katex=display] W_{rocket} = KE + PE_g + W_{air} [/katex].

In plain english this would read: “The work done by the rocket transforms into some kinetic energy, some potential energy, and some work done by air resistance.”

From here, substitute in the equations for each type of energy. Then solve for [katex] W_{air} [/katex].

Practice it

Now its your turn. Try the 4 questions below. For more difficulty levels you can use UBQ to sort through even more energy questions.

Open in UBQ
Question 2
Intermediate
Proportional Analysis

A \(4.0 \, \text{kg}\) block is moving at \(5.0 \, \text{m/s}\) along a horizontal frictionless surface toward an ideal spring that is attached to a wall. After the block collides with the spring, the spring is compressed a maximum distance of \(0.68 \, \text{m}\). What is the speed of the block when the spring is compressed to only one-half of the maximum distance?

Open in UBQ
Question 3
Intermediate
Mathematical

A projectile of mass 0.750 kg is shot straight up with an initial speed of 18.0 m/s.

Part (a)3 pts

How high would it go if there no air resistance?

Part (b)3 pts

If the projectile rises to a maximum height of only 11.8 m, determine the magnitude of the average force due to air resistance.

Part (c)3 pts

If the speed of the projectile is doubled in part a, by what factor would the height change by? Justify.

Open in UBQ
Question 4
Intermediate
Mathematical

A 84.4 kg climber is scaling the vertical wall. His safety rope is made of a material that behaves like a spring  that has a spring constant of 1.34 x 103 N/m. He accidentally slips and falls 0.627 m before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?

Open in UBQ
Question 5
Advanced
Mathematical
A horizontal force of \(110 \, \text{N}\) is applied to a \(12 \, \text{kg}\) object, moving it \(6 \, \text{m}\) on a horizontal surface where the kinetic friction coefficient is \(\mu_k = 0.25\). The object then slides up a \(17^\circ\) inclined plane. Assuming the \(110 \, \text{N}\) force is no longer acting on the incline, and the coefficient of kinetic friction there is \(\mu_k = 0.45\), calculate the distance the object will slide on the incline.

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KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

Metric Prefixes

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]

  3. Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]

  4. Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

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