# Energy, Work, and Power Speed Review

###### Jason Kuma

Writer | Coach | Builder | Fremont, CA

###### Article Content
Energy is a fundamental concept in physics that is involved in a wide range of phenomena. It is the ability to do work or produce change, and exists in various forms such as kinetic energy, potential energy, thermal energy, and electrical energy. In this article, we will explore these different forms of energy and how to solve energy problems.

### Forms of Energy

There are 3 main types of energy.

• Kinetic (any mass that has velocity)
• Potential (any mass has stored energy)
• Work (all other types of energy that isn’t potential or kinetic).

Note that Elastic Energy also called spring potential energy is a type of potential energy.

Note that Mechanical Energy is only the sum of Kinetic and Potential energy.

Now lets dive into each type of energy and cover some misconceptions.

#### Kinetic Energy

Kinetic energy (KE) is the energy of motion, and is given by the formula:

[katex] KE = \frac{1}{2}mv_2 [/katex]

For example, a car moving at a high speed has more kinetic energy than a car moving at a slow speed.

Now to speed something up would mean to increase kinetic energy, which would come from an external source (Work energy). This is why [katex] \Delta KE = \text{Work} [/katex]

#### Potential Energy

Potential energy is the energy of position or configuration. It is stored energy that an object possesses due to its position or arrangement. There are several types of potential energy, such as gravitational potential energy, elastic potential energy, and chemical potential energy.

Gravitational potential energy (PEg) is the energy of an object due to its height. It is given by the formula

[katex] PE_g = mgh [/katex]

Where “m” is mass, “g” is the value of gravity, and “h” is the height of the object.

For example, a rock at the top of a hill has more gravitational potential energy than a rock at the bottom of the hill.

Elastic (or Spring) potential energy is the energy stored in a stretched or compressed spring. It is given by the formula

[katex] El = \frac{1}{2}kx^2 [/katex]

Where “k” is the spring constant (how stiff the spring is), and “x” is the amount of stretch/compression in meters.

For example, a spring that is stretched more has more elastic potential energy than a spring that is stretched less.

Chemical potential energy is the energy stored in the chemical bonds of a substance. It is the energy that is released or absorbed during a chemical reaction. For example, gasoline has a high chemical potential energy, which is released when it is burned to produce energy.

#### Work Energy

[katex] \text{Work} = Fd [/katex]

In simple terms work is the force applied over a certain distance. Pushing a box is a simple example of positive work being done. The force of friction is an example of negative work being done.

Keep in mind the other definition of work:

[katex] \text{Work} = \Delta KE [/katex]

There are two important things to note. (1) Energy is a scalar. Positive and negative simply mean that the object is either gaining (+) or loosing (-) energy. (2) Force and displacement MUST be parallel to each other in order for there to be work energy.

#### Other Types of Energy

Here are a few more types of energy, not covered in the AP Physics 1 exam.

Thermal energy is the energy of heat. A common example is friction. It is the energy that is transferred from one body to another as a result of a temperature difference. Thermal energy is related to the temperature of an object, with hotter objects having more thermal energy than cooler objects.

Electrical energy is the energy of electric charge. Energy is transferred through an electric circuit due to the force applied over a distance (Work done) on charged particles.

### Energy Conservation

The law of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. This means that the total amount of energy in a closed system remains constant, even if the energy changes form.

No change in energy → [katex] \Delta E = 0 \rightarrow E_f – E_i = 0 \rightarrow E_f = E_i [/katex]. In simple terms, this means the amount of energy a system starts with should be the exact same amount it ends with.

For example, consider a ball being thrown into the air. The ball has kinetic energy due to its motion, and it also has gravitational potential energy due to its height. As the ball rises, its kinetic energy decreases and its gravitational potential energy increases, but the total amount of energy remains constant.

### Framework for Solving Energy Problems

When solving energy problems, it is important to identify the forms of energy involved and use the appropriate formulas to calculate the energy.

1. Start with the Law of Conservation of Energy which states → [katex] E_i = E_f [/katex]
2. Read the scenario and identify the initial forms of energy. Write it down under [katex] E_i [/katex]
3. Identify the final forms of energy. Write it down under [katex] E_f [/katex]
4. Substitute the equations for each form of energy. Rearrange your equation and solve for asked variable.

Let’s practice this framework with 3 commonly asked AP Physics 1 questions on Energy

#### Example Problem #1

A ball is thrown straight up into the air with an initial velocity of 10 m/s. What is the maximum height reached by the ball?

Show Further Explanation

The maximum height of the ball is determined by the balance between its initial kinetic energy and the gravitational potential energy it gains as it rises. As the ball rises, its kinetic energy decreases and its gravitational potential energy increases, until they are equal at the maximum height. At this point, the ball stops rising and begins to fall back down.

#### Example Problem #2

A 0.5 kg block is pushed up a frictionless ramp with a constant force of 5 N. The block starts at rest at the bottom of the ramp and reaches a height of 2 meters at the top. How much work is done on the block by the applied force?

Show Further Explanation

As the block is pushed up the ramp, it gains gravitational potential energy due to its increasing height. The work done on the block by the applied force is equal to the change in the block’s gravitational potential energy, which is equal to the applied force times the distance the block moves.

*You might wonder why we couldn’t simply use W = Fd. It’s because we are given the height NOT the length (d) of the ramp. Remember that the distance and force applied must be parallel.

### Conservative vs Non-Conservative Systems

Let’s clear up a misconception: “Non-conservative” does NOT mean energy is not conserved in the system. [katex] E_i = E_f [/katex] (Energy is conserved) as long as no net external forces act on the system.

A conservative force does not depend on the path an object takes. Think about gravity acting on a block. Wether it’s sliding down a ramp, or being dropped straight down from the height of the ramp, it will still have the same energy.

A non-conservative force does depend on the path taken. Think about friction. If you push a block up or down a ramp, there will be energy “dissipated” that can’t be transformed back into potential or kinetic energy.

Important things to remember:

• Work done by ALL non-conservative forces, is just [katex] \text{Work} = Fd [/katex]
• Another way to way to find the work done by a non-conservative force → [katex] W_{NC} = \Delta KE + \Delta PE [/katex]. This simply tells us that the amount of energy lost or gained from both potential and kinetic energies is the total amount of work done by a non-conservative force.
• If there is NO work done by non-conservative forces in a system, then the system has purely mechanical energy.
• HINT: use the problem solving framework above to solve any energy question, regardless of the types of forces in the system.

### Power

Power is a measure of the amount of energy dissipated over a period of time.

[katex] P = \frac{W}{t} [/katex]

Where “W” is the work done and “t” is the time taken.

The units are Joules/second. We commonly call this Watts (common in light bulbs) or Horsepower (common in cars).

### Master Energy Even Quicker

If any of this sounds confusing, you’re not alone! I’ve helped 100’s of Physics students to achieve the highest marks on their in class exams and on the AP Physics 1 test.

There are 1-to-1 private lessons OR group lessons available for all students who want to get ahead, make physics stupidly simple, and get the highest grades. Click here to view our programs.

### Try these 10 Energy Questions

Question 1
Difficulty - Intermediate
Solve Type - Mathematical
A 0.5 kg cart, on a frictionless 2 m long table, is being pulled by a 0.1 kg mass connected by a string and hanging over a pulley. The system is released from rest. After the hanging mass falls 0.5 m, calculate the speed of the cart on the table. Use ONLY forces and energy.
Question 2
Difficulty - Intermediate
Solve Type - Conceptual
Two balls are thrown off a building with the same speed, one straight up and one at a 45° angle. Which statement is true if air resistance can be ignored?
1. Both hit the ground at the same time.

2. Both hit the ground with the same speed.

3. The one thrown at an angle hits the ground with a lower speed.

4. The one thrown at an angle hits the ground with a higher speed.

5. Both (a) and (b)

Question 3
Difficulty - Intermediate
Solve Type - Conceptual
A satellite in circular orbit around the Earth moves at constant speed. This orbit is maintained by the force of gravity between the Earth and the satellite, yet no work is done on the satellite. How is this possible?
Question 4
Difficulty - Intermediate
Solve Type - Mathematical
A child pushes horizontally on a box of mass m with constant speed v across a rough horizontal floor. The coefficient of friction between the box and the floor is µ. At what rate does the child do work on the box?
1. [katex] \mu mgv [/katex]

2. [katex] \mu mg/v  [/katex]

3. [katex]  \mu mg/v^2[/katex]

4. [katex] mgv [/katex]

5. Can not be determined.

Question 5
Difficulty - Intermediate
Solve Type - Conceptual
An object is projected vertically upward from ground level. It rises to a maximum height [katex] H [/katex]. If air resistance is negligible, which of the following must be true for the object when it is at a height [katex] H/2 [/katex] ?
1. Its potential energy is half of its initial potential energy.

2. Its speed is half of its initial speed.

3. Its total mechanical energy is half of its initial value.

4. Its kinetic energy is half of its initial kinetic energy.

5. None of the above.

Question 6
Difficulty - Intermediate
Solve Type - Mathematical
A person is making homemade ice cream. She exerts a force of magnitude 23 N on the free end of the crank handle on the ice-cream maker, and this end moves on a circular path of radius 0.25 m. The force is always applied parallel to the motion of the handle. If the handle is turned once every 1.7 s, what is the average power being expended?
Question 7
Solve Type - Conceptual
In which one of the following circumstances does the principle of conservation of mechanical energy apply, even though a nonconservative force acts on the moving object?
1. The nonconservative force has a component that points opposite to the displacement of the object.

2. The nonconservative force is perpendicular to the displacement of the object.

3. The nonconservative force has a direction that is opposite to the displacement of the object.

4. The nonconservative force has a component that points in the same direction as the displacement of the object.

5. The nonconservative force points in the same direction as the displacement of the object.

Question 8
Solve Type - Mathematical
A 100 kg person is riding a 10 kg bicycle up a 25° hill. The hill is long and the coefficient of static friction is 0.9. The person rides 10 m up the hill then takes a rest at the top. If she then starts from rest from the top of the hill and rolls down a distance of 7 m before squeezing hard on the brakes locking the wheels. How much work is done by friction to bring the bicycle to a full stop, knowing that the coefficient of kinetic friction is 0.65?
Question 9
Solve Type - Conceptual
A lighter car and a heavier truck, each traveling to the right with the same speed [katex] v [/katex] hit their brakes. The retarding frictional force F on both cars turns out to be constant and the same. After both vehicles travel a distance [katex] D [/katex] (and both are still moving), which of the following statements is true?
1. They will traverse the distanced in the same time.

2. They will have the same velocity.

3. The work done on both vehicles is the same.

4. The average power expended is the same.

5. They will have the same kinetic energy

Question 10
Solve Type - Mathematical
One end of a spring is attached to a solid wall while the other end just reaches to the edge of a horizontal, frictionless tabletop, which is a distance [katex] h [/katex] above the floor. A block of mass M is placed against the end of the spring and pushed toward the wall until the spring has been compressed a distance [katex] x [/katex]. The block is released and strikes the floor a horizontal distance [katex] D [/katex] from the edge of the table. Air resistance is negligible. Derive an expressions for the following quantities only in terms of [katex] M, x, D, h, [/katex] and any constants.
###### 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
[katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex][katex]F = ma[/katex]
[katex]v = v_i + at[/katex][katex]F_g = \frac{G m_1m_2}{r^2}[/katex]
[katex]a = \frac{\Delta v}{\Delta t}[/katex][katex]f = \mu N[/katex]
[katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex]
Circular MotionEnergy
[katex]F_c = \frac{mv^2}{r}[/katex][katex]KE = \frac{1}{2} mv^2[/katex]
[katex]a_c = \frac{v^2}{r}[/katex][katex]PE = mgh[/katex]
[katex]KE_i + PE_i = KE_f + PE_f[/katex]
MomentumTorque and Rotations
[katex]p = m v[/katex][katex]\tau = r \cdot F \cdot \sin(\theta)[/katex]
[katex]J = \Delta p[/katex][katex]I = \sum mr^2[/katex]
[katex]p_i = p_f[/katex][katex]L = I \cdot \omega[/katex]
Simple Harmonic Motion
[katex]F = -k x[/katex]
[katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex]
[katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex]
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]\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]

General Metric Conversion Chart

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]

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 [katex] v_i [/katex] is written as [katex] u [/katex]; sometimes [katex] \Delta x [/katex] is written as [katex] s [/katex].
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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