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.


Work measures a change in energy caused by a force.
Work is only done when a force causes an object to change position.

d Θ F

$$W = Fd\cos\theta$$

\(W\) = 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.

5 kg d = 200 m Θ = 30° F = 100 N 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?

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

m v

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

\(K\) = 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:

  • gravitational potential energy from being high up.
  • a compressed spring
  • two opposing magnets pushed together
  • chemical energy in a battery or gasoline
  • 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.
    m h

    $$U_{g} = mgh$$

    \(U_g\) = 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?

    5 kg y = 100 m y = 0 m g = 9.8 m/s²

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

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

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

    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.

    100 kg y = 15 m y = 0 g = 9.8 m/s²
    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 $$\text{6th story - 1st story = 5 stories} $$ $$ \left(5\,\mathrm{story} \right) \left(\frac{3 \, \mathrm{m}} {1 \, \mathrm{story}}\right) = 15 \, \mathrm{m}$$
    $$U_{g} = mgh$$ $$U_{g} = (100)(9.8)(15)$$ $$U_{g} = 14700 \, \mathrm{J}$$
    80 kg y = 0 y = -400 ft g = 9.8 m/s²
    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:

    x k F

    $$F = -kx$$

    \(F\) = 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 $$200 \, \mathrm{(c)m} = \mathrm{200\left(\frac{1}{100}\right)m = 2\,m}$$
    $$U_{s} = \tfrac{1}{2}kx^{2}$$ $$U_{s} = \tfrac{1}{2} (0.3)(2)^{2}$$ $$U_{s} = 0.6 \, \mathrm{J}$$
    k=? F = 100 N x = 40 cm
    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 $$x= \mathrm{40\,(c)m = 40\left(\frac{1}{100}\right)m =0.4 \, m}$$
    $$F= -kx$$ $$-\frac{F}{x} = k$$ $$-\frac{-100}{0.4} = k$$ $$250 \mathrm{\tfrac{kg}{s^2}} = k$$