# How to Solve Any AP Physics Energy Question

###### Jason Kuma

Writer | Coach | Builder | Fremont, CA

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

AP Physics covers only 3 types of energy:

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

Lastly every type of energy thats not the 3 types above is classified as Work Energy where W = Fd . 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: E_A = E_B , 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.

Solve Type - Mathematical
A 2 kg model rocket is launched with a thrust force of 275 N and reaches a height of 90 m, moving at 150 m/s at its peak. 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:

E_A = E_B

W_{rocket} = KE + PE_g + W_{air} .

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 W_{air} .

### 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.

Question 1
Difficulty - Intermediate
Solve Type - Mathematical

A 4.0-kg block is moving at 5.0 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 m. What is the speed of the block when the spring is compressed to only one-half of the maximum distance?

Question 2
Difficulty - Intermediate
Solve Type - Mathematical

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

Question 3
Difficulty - Intermediate
Solve Type - 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?

Question 4
Solve Type - Mathematical
A horizontal force of 110 N is applied to a 12 kg object, moving it 6 m on a horizontal surface where the kinetic friction coefficient is 0.25. The object then slides up a 17° inclined plane. Assuming the 110 N force is no longer acting on the incline, and the coefficient of kinetic friction there is 0.45, calculate the distance the object will slide on the incline.
###### Jason Kuma

Writer | Coach | Builder | Fremont, CA

## Units in AP Physics 1

Unit 1 – Linear Kinematics

Unit 2 – Linear Forces

Unit 3 – Circular Motion

Unit 4 – Energy

Unit 5 – Momentum

Unit 6 – Torque

Unit 7 – Oscillations

Unit 8 – Fluids

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KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g}
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

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

1. Start with the given measurement: \text{5 km}

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

3. Perform the multiplication: \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}

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

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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