# 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.

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

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

$$K$$ = 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:

• a battery stores electrical energy
• being up high is gravitational energy
• a compressed spring
• two opposing magnets pushed together
• chemical energy, like gasoline or wood.

# 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?

 5kg y = 100m y = 0m g = 9.8m/s² $$U_{g} = mgh$$ $$U_{g} = (5)(9.8)(100)$$ $$U_{g} = 4900J$$ 5kg y = 0 y = -100m g = 9.8m/s² $$U_{g} = mgh$$ $$U_{g} = (5)(9.8)(0)$$ $$U_{g} = 0J$$ 5kg y = 50m y = 0 y = -50m g = 9.8m/s² $$U_{g} = mgh$$ $$U_{g} = (5)(9.8)(50)$$ $$U_{g} = 2450J$$

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
 $$\left(6 story-1 story \right) \left(\frac{3m}{1story}\right) = 15m$$ $$U_{g} = mgh$$ $$U_{g} = (100)(9.8)(15)$$ $$U_{g} = 14700J$$ 100kg y = 15 y = 0m g = 9.8m/s²
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
 $$-400ft\left(\frac{0.3048m}{1ft}\right) = -122m$$ $$U_{g} = mgh$$ $$U_{g} = (80.0)(9.8)(-122)$$ $$U_{g} = -95648J$$ 80kg y = 0 y = -122m g = 9.8m/s²

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