The best way to understand energy is to learn about its different forms.

Energy changes forms without loss or gain. 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 cannot be created or destroyed, but it can change forms.

- Chemical energy is converted to kinetic energy in the explosion of a stick of dynamite.
- A microwave converts electrical energy into light energy and then into the heat energy.
- As an object falls the gravitational potential energy of position is converted to kinetic energy.
- Chemical energy stored in methane is burned to form heat energy.

# Work

Work measures a change in energy caused by a force.

**Work is only done when a force causes an object to change position.**

- An elevator lifting an person is work.
- Pedaling a moving bike is work.
- Pushing a box across the floor is work.
- A car skidding to a stop is work.
- Pushing a stationary wall is not work.
- A ball rolling on flat frictionless ground is not 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** returns the component of the force vector in the same direction as the displacement vector.
vector.

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

## solution

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

## solution

$$W = Fd\cos\theta$$ $$W = (10)(2)\cos(0)$$ $$W = (10)(2)(1)$$ $$W = 20\, \mathrm{J}$$**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.

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

# Kinetic Energy

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

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

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

## derivation of kinetic energy

$$v^2 = u^2 +2a\Delta x$$

Kinetic energy is defined as relative to zero velocity, so set the initial velocity to zero.

$$v^2 = 2a\Delta x$$ $$a\Delta x =\frac{v^2}{2}$$

times both sides by mass

$$ma\Delta x =\tfrac{1}{2}mv^2 $$ $$F=ma$$ $$F\Delta x =\tfrac{1}{2}mv^2 $$Kinetic energy can be produced by work.

$$K = W = Fd = F\Delta x$$ $$K =\tfrac{1}{2}mv^2$$Kinetic energy is the energy objects have from motion. This includes anything moving:
*light, sound, heat, electricity*

Energy has units of Joules in the metric system. 1 Joule is the same as a 2 kg mass moving at 1 m/s.

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.

**Example:**What is the kinetic energy of Zoe running at 5 m/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

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

**left**?

## solution

$$K = \tfrac{1}{2}mv^{2}$$ $$K = \tfrac{1}{2}(20)(-5)^{2}$$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 when she runs left or right.

$$K = 250 \, \mathrm{J}$$**Example:**Let's imagine what 100 J feels like. How fast would an 80 kg person have to be moving to have 100 J?

## solution

$$K = \tfrac{1}{2}mv^{2}$$ $$100 = \tfrac{1}{2}(80)v^{2}$$ $$100 = (40)v^{2}$$ $$2.5 = v^{2}$$ $$1.58 \mathrm{\tfrac{m}{s}} = v$$That's about walking speed. It takes about 100 J to accelerate a person from rest to walking speed.

# Potential Energy

Potential energy is stored energy from the position of an object.

Examples of potential energy:

How could each of these potential energies convert into kinetic energy?

# 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, 9.8 on Earth [m/s²]

\(h\) = 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?

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

We get 3 different answers because of the different zero points. Unlike kinetic energy, gravitational potential energy
isn't consistent, 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 100 kg person from the 1st story to the 6th story? convert 1 story to meters

## solution

$$\text{6th story - 1st story = 5 stories} $$ $$ \left(5\,\mathrm{story} \right) \left(\frac{3 \, \mathrm{m}} {1 \, \mathrm{story}}\right) = 15 \, \mathrm{m}$$$$U_{g} = mgh$$ $$U_{g} = (100)(9.8)(15)$$ $$U_{g} = 14700 \, \mathrm{J}$$

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

*LEX LUTHOR: Drop of Doom*?

convert ft to m

## solution

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

# 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. $$W = \int_{0}^{x} kx \ dx$$ $$W = \frac{1}{2}
kx^2 $$

# $$U_{s} = \tfrac{1}{2}kx^{2}$$

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

\(x\) = displacement [m]

k = m =

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

## solution

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

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

## solution

$$x= \mathrm{40\,(c)m = 40\left(\frac{1}{100}\right)m =0.4 \, m}$$$$F= -kx$$ $$-\frac{F}{x} = k$$ $$-\frac{-100}{0.4} = k$$ $$250 \mathrm{\tfrac{kg}{s^2}} = k$$