# Conservation of Energy

Take a look at the simulated spring and mass system below.

k =     m =

What types of energy are being traded?

When is the velocity the highest?

What is the effect of a higher spring constant?

k = m =

This spring simulation is aligned vertically. It has gravitational potential energy in addition to kinetic and spring energy.

When is kinetic energy the highest?

When is gravitational energy the lowest?

When is spring energy the lowest?

What is the effect of a lower mass?

Why does the total energy stay the same?

# Conservation of Energy

Energy is conserved. This means it can't be reduced or increased, but as systems change, the energy changes form. Since total energy is always the same value, we can build an equation that sets the total energy at one time equal to the total energy at any other time.

Energy cannot be created or destroyed, but it can change forms.

# $$E_i=E_f$$

$$E_i$$ = initial energy[J]
$$E_f$$ = final energy [J]

This equation is bare bones right now. We need to change it for each situation. We should only include energy types that are changed. For now we can calculate three types of energy.

• If an object changes speed, include kinetic energy.
• If an object moves vertically, include gravitational potential energy.
• If a spring is compressed, include spring potential energy.
• For example, if a ball at rest starts high up and falls down, the conservation of energy equation should only include kinetic and gravitational energy.

$$K_i + U_i = K_f + U_f$$ $$\frac{1}{2}mu^2 + mgh_i = \frac{1}{2}mv^2 + mgh_f$$

We can often set some the initial or final energies to zero as well. Initial kinetic energy is zero because the ball starts at rest. Choosing the final height to be zero will set the final gravitational potential energy to zero.

$$mgh_i = \frac{1}{2}mv^2$$

Also, look for variables that show up in every term. In this case mass can be ignored by dividing everything by mass.

$$gh_i = \frac{1}{2}v^2$$
Example: A ball falls from rest off a 0.80 m table. How fast is the ball going just before it hits the ground?
solution $$U_g = K$$ $$mgh_{i} = \tfrac{1}{2}mv^{2}$$ $$\text{m is in every term.}$$ $$\frac{mgh_{i}}{m} = \frac{\tfrac{1}{2}mv^{2}}{m}$$ $$gh_{i} = \tfrac{1}{2}v^{2}$$ $$(9.8)(0.80) = \tfrac{1}{2}v^{2}$$ $$15.68 = v^{2}$$ $$\pm 3.959 \mathrm{ \tfrac{m}{s} } = v$$
Example: If you throw a half-filled water bottle straight up at 10 m/s, how high will it go?
solution $$K = U_g$$ $$\tfrac{1}{2}mv^{2} = mgh$$ $$\tfrac{1}{2}v^{2} = gh$$ $$\tfrac{1}{2}(10)^{2} = 9.8h$$ $$50 = 9.8h$$ $$5.10 \, \mathrm{m} = h$$
Example: A 0.43 kg soccer ball kicked at 10 m/s rolls down a 30 m tall hill. How fast will the ball be moving at the bottom of the hill?
solution $$U_{i} + K_{i} = U_{f} + K_{f}$$ $$mgh_{i} + \tfrac{1}{2}mu^{2} = mgh_{f} +\tfrac{1}{2}mv^{2}$$ $$gh_{i} + \tfrac{1}{2}u^{2} = gh_{f} +\tfrac{1}{2}v^{2}$$ $$(9.8)(30) + \frac{1}{2}(10)^{2} = (9.8)(0) + \frac{1}{2}v^{2}$$ $$294 + 50 = \tfrac{1}{2}v^{2}$$ $$344 = \tfrac{1}{2}v^{2}$$ $$\pm 26.2 \mathrm{\tfrac{m}{s}} = v$$
Example: A 0.20 kg ball is placed on a spring compressed down 0.15 m. The internet says that your spring has a spring constant of 200.0 kg/s². How high could the ball go just using the energy stored in the spring?
solution $$U_{s} = U_{g}$$ $$\tfrac{1}{2}kx^{2} = mgh$$ $$\text{Can't divide by mass}$$ $$\tfrac{1}{2}(200)(0.15)^{2} = (0.2)(9.8)h$$ $$1.15 \, \mathrm{m} = h$$
Example: A roller coaster cart starts at rest 126 m high on the top of the first drop on "Superman: Escape from Krypton" at Six Flags Magic Mountain. How fast is the cart going at the bottom of the hill?
solution $$U_{i} + K_{i} = U_{f} + K_{f}$$ $$mgh_{i} + \tfrac{1}{2}mu^{2} = mgh_{f} +\tfrac{1}{2}mv^{2}$$ $$gh_{i} + \tfrac{1}{2}u^{2} = gh_{f} +\tfrac{1}{2}v^{2}$$ $$(9.8)(126) + \tfrac{1}{2}(0)^{2} = (9.8)(0) +\tfrac{1}{2}v^{2}$$ $$1234.8 = \tfrac{1}{2}v^{2}$$ $$\pm 49.69 \, \mathrm{\tfrac{m}{s}} = v$$
Example: A 0.20 kg ball is placed on a spring compressed down 0.15 m. The internet says that your spring has a spring constant of 200.0 kg/s². How fast should the ball be moving right after it leaves the spring?
solution $$U_{s} = K$$ $$\tfrac{1}{2}kx^{2} = \tfrac{1}{2}mv^{2}$$ $$\tfrac{1}{2}(200)(0.15)^{2} = \tfrac{1}{2}(0.2)v^{2}$$ $$22.5 = v^{2}$$ $$\pm 4.74 \, \mathrm{\tfrac{m}{s}} = v$$