# What is Energy?

Energy is such a fundamental concept that there isn't a simple definition. Lets look at some of it's different forms.

wiki: Forms of energy

Energy can change forms without loss or gain. In physics, the law of **conservation of energy** states
that the total energy of an isolated system remains constant—it is said to be conserved over time.

### Energy can be neither created nor be destroyed, but it transforms from one form to another.

**Example:** Chemical energy is converted to kinetic energy in the explosion of a stick of dynamite.

**Example:** As an object falls the gravitation potential energy of position is converted to kinetic
energy.

**Example:** A microwave converts electical energy into Electro-magnetic microwave energy and then into
the energy of heat as food heats up.

**Example:** Chemical energy stored in methane is burned to form heat energy. That heat energy expands
and rises up to pass through a fan. The fan spins around to make rotational kinetic energy. That energy is turned
into electrical energy because of magnets pulling on the electrons in a wire. This makes electrical energy. That
energy flows down the wire, ultimatly powering my computer, which turns that enegy into heat.

# Work

Work can measure a change in energy caused by a force. **Work is done when a force causes an object to displace.**

**Examples:**

An elevator lifting an object up several floors is work.

Electrical energy is converted to gravitational potential energy.

You pushing a box across a frictionless surface is work.

Chemical energy in your muscles makes kinetic energy.

A car skidding to a stop is work.

Kinetic energy is converted to mostly heat and some sound energy.

Pushing a stationary wall isn't work.

A ball rolling on frictionless level ground isn't work.

# $$W = Fd\cos\theta$$

\(W\) = work [J, Joules, kg m²/s²]\(F\) = Force [N, Newtons, kg m/s²]

*vector*

\(d\) = displacement [m]

*vector*

\(\theta\) = angle between applied force and displacement

**F cos θ** gives us only the component of the force that is in the same direction as the displacement.
Remember vectors?

**Example:**If I apply 100N of force at an angle 30 degrees above the horizontal and I move a 5kg box 200 meters horizontally, how much work is done?

## solution

$$W = Fd\cos\theta$$ $$W = (100)(200)\cos(30)$$ $$W = (100)(200)(0.866)$$ $$W = 17320J$$**Example:**If I apply 10N of horizontal force to move a dry erase marker 2 meters horizontally, how much work is done?

## solution

$$W = Fd\cos\theta$$ $$W = (10)(2)\cos(0)$$ $$W = (10)(2)(1)$$ $$W = 20J$$**Example:**What is the least amount of work it would take to lift a Tesla (2,085 kg) up 2.5 meters?

## solution

The force to lift must be slightly larger than the force of gravity.

$$F_{g} = mg$$$$W = Fd\cos\theta$$ $$W = (mg)d\cos\theta$$ $$W = (2085)(9.8)(2.5)\cos(0)$$ $$W = 51082.5J$$

# Kinetic Energy

# $$K = \small\frac{1}{2}\normalsize mv^{2}$$

\(K\) = kinetic energy [J, Joules]\(m\) = mass [kg]

\(v\) = velocity [m/s]

Kinetic energy is the energy massive objects have from motion.

Energy has units of Joules in the metric system. 1 Joule is the energy of a 2kg mass moving at 1m/s. Other units of energy are: kilowatt-hour, Btu, pounds of TNT, and calorie.

Energy is a **scalar**. It has no direction. Not having to work with vectors can make solving 2-D and
3-D problems much easier than with the equations of motion.

**Example:**What is the kinetic energy of Zoe running at 5m/s to the right? (Zoe is a dog. She has a mass of 20kg. She is 0.56 meters tall. She has reddish fur, like a fox.)

## solution

$$KE = \frac{1}{2}mv^{2}$$ $$K = \frac{1}{2}(20)(5)^{2}$$ $$K = 250J$$**Example:**What is the kinetic energy of Zoe running at 5m/s to the

**left**?

## solution

$$K = \frac{1}{2}mv^{2}$$ $$K = \frac{1}{2}(20)(-5)^{2}$$Energy is a scalar so direction doesn't matter. The negative sign on the velocity goes away because it is squared, and we get the same answer.

$$K = 250J$$# Potential Energy

Potential energy is energy that could potentially do some work.

Examples of potential energy are:

# Gravitational Potential Energy

If you drop a massive object it will gain kinetic energy as it falls. We call the potential for the energy of its position to turn into kinetic energy gravitational potential energy

## derivation of gravitational potential energy

Gravitational energy can be calculated with the work equation. $$W = Fd\cos\theta$$ As an object falls the force of gravity pulls the object vertically down. $$F_{g} = mg$$ $$W = mgd\cos\theta$$ The displacement is vertical so lets call it height: h. $$W = mgh\cos\theta$$ The force of gravity and the height are pointed in the same direction, so cos(0)=1. $$W = mgh$$ We now have a nice equation for the energy that comes from the work of lifting an object.

# $$U_{g} = mgh$$

\(U_g\) = gravitational potential energy [J, Joules]\(m\) = mass [kg]

\(g\) = acceleration from gravity (on earth = 9.8) [m/s²]

*vector*

\(h\) = height [m]

**Height** is the tricky part of this equation, because where is zero height? The ground floor can be
the zero point, or the floor of the basement or the top of a table, or any point. You choose a zero point that
makes the problem less complex, like the starting or ending point.

Which of these three zero points seem the best choice?

$$U_{g} = mgh$$ $$U_{g} = (5)(9.8)(100)$$ $$U_{g} = 4900J$$ |
$$U_{g} = mgh$$ $$U_{g} = (5)(9.8)(0)$$ $$U_{g} = 0J$$ |
$$U_{g} = mgh$$ $$U_{g} = (5)(9.8)(50)$$ $$U_{g} = 2450J$$ |

We get 3 different answers because of the different zero points. Unlike kinetic energy, gravitational potential energy
isn't consistant, but we can still use it to find a **change in gravitational potential energy**.

**Example:**How much energy does it take for an elevator to bring a 100kg person from the 1st story to the 6th story? convert 1 story to meters

## solution

$$ \left(6 story-1 story \right) \left(\frac{3m}{1story}\right) = 15m$$ $$U_{g} = mgh$$ $$U_{g} = (100)(9.8)(15)$$ $$U_{g} = 14700J$$ |

**Example:**How much gravitational energy is lost as a 80.0kg person rides down from the top of the 400ft ride

*LEX LUTHOR: Drop of Doom*? convert ft to m

## solution

$$-400ft\left(\frac{0.3048m}{1ft}\right) = -122m$$ $$U_{g} = mgh$$ $$U_{g} = (80.0)(9.8)(-122)$$ $$U_{g} = -95648J$$ |

# Elastic Potential Energy

Elastic materials all obey **Hooke's law** which says that there is a linear relationship between displacement
and the restoring force:

# $$F = -kx$$

\(F\) = Restoring elastic force [N, Newtons, kg m/s²]*Vector*

\(k\) = spring constant [kg/s²]

\(x\) = displacement from equilibrium [m]

*vector*

The spring constant k is different for every spring. Sometimes when you buy a spring the constant will be listed, but we can also measure k directly.

The Spring force is always directed back towards the equilibrium point. This works for compressing the spring or expanding the spring.

## derivation of elastic potential energy

The energy stored in an elastic material can be calculated with the equation for work and some calculus. $$W = Fd\cos\theta$$
The displacement in the work equation d is the same as the displacement in the force equation x. Lets make
them both x. $$W = Fx\cos\theta$$ The elastic force and the displacement are always at 0 degrees. cos(0)
= 1 $$W = Fx$$ The force is the elastic force. $$F = -kx$$ We can't directly replace F with -kx in the work
equation because the force changes over the displacement.

F= 0 at x=0 and F=-k*10 at x=10

We need to use an operation from calculus called an integral.

The integral will find the area of F times x, which equals work. $$W = \int_{0}^{x} kx \ dx$$ $$W = \frac{1}{2}
kx^2 $$

# $$U_{s} = \small\frac{1}{2}\normalsize kx^{2}$$

\(U_s\) = elastic/spring potential energy [J, Joules, kg m²/s²]\(k\) = spring constant [kg/s²]

\(x^2\) = displacement [m²]

**Example:**Find the elastic energy of a 500g cylinder stuck to a spring (k = 0.3) that is 200cm away from its equilibrium point?

## solution

$$200(c)m = 200\left(\frac{1}{100}\right)m = 2m$$$$U_{s} = \small\frac{1}{2}\normalsize kx^{2}$$ $$U_{s} = \small\frac{1}{2}\normalsize (0.3)(2)^{2}$$ $$U_{s} = 0.6J$$

**Example:**You need to find the spring constant for a spring mass system. To find the spring constant you pull the mass 40cm out of equilibrium and measure a spring force of 100N. What's the spring constant?

## solution

$$40(c)m = 40\left(\frac{1}{100}\right)m =0.4m$$$$F= -kx$$ $$-\frac{F}{x} = k$$ $$-\frac{-100}{0.4} = k$$ $$250 = k$$