Conservation of Energy

Trading Energy

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

What types of energy are being traded?

When is the velocity the highest?

What is the effect of a higher spring constant?


 k =     m =





  k = m =

This simulation is a bit more fun with gravitational potential energy as well.

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?

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.

$$K_i+U_i = K_f+U_f$$

\( K \) = kinetic energy [J]
\( U \) = potential energy [J]

This equation is barebones right now. We need to change it for each situation.

We should only include energy types that are changed.
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.

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

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.

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

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

$$gh_i = \frac{1}{2}v^2_f$$
Example: A ball falls off a table from 0.80m high. How fast is the ball going just before it hits the ground?
solution $$U_g = K$$ $$mgh_{i} = \frac{1}{2}mv^{2}$$

m is in every term. Let's divide by m.

$$ \frac{mgh_{i}}{m} = \frac{\frac{1}{2}mv^{2}}{m}$$ $$gh_{i} = \frac{1}{2}v^{2}$$ $$(9.8)(0.80) = \frac{1}{2}v^{2}$$ $$15.68 = v^{2}$$ $$\pm 3.959\small\frac{m}{s} =\normalsize v$$
Example: If you throw a half-filled water bottle straight up at 10 m/s, how high will it go?
solution $$K = U_g$$ $$\frac{1}{2}mv^{2} = mgh$$

m is in every term. Let's divide by m.

$$\frac{1}{2}v^{2} = gh$$ $$\frac{1}{2}(10)^{2} = (9.8)h$$ $$50 = (9.8)h$$ $$5.10m = h$$
Example: A roller coaster cart starts at rest 126m 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 $$U_{i} + K_{i} = U_{f} + K_{f}$$ $$mgh_{i} + \frac{1}{2}mu^{2} = mgh_{f} +\frac{1}{2}mv^{2}$$

m is in every term. Let's divide by m.

$$gh_{i} + \frac{1}{2}u^{2} = gh_{f} +\frac{1}{2}v^{2}$$ $$(9.8)(126) + \frac{1}{2}(0)^{2} = (9.8)(0) +\frac{1}{2}v^{2}$$ $$1234.8 = \frac{1}{2}v^{2}$$ $$\pm 49.69 \small\frac{m}{s} =\normalsize v$$
Example: A 0.43kg soccer ball kicked at 10m/s rolls down a 30m tall hill. How fast will the ball be moving at the bottom of the hill?
solution $$U_{i} + K_{i} = U_{f} + K_{f}$$ $$mgh_{i} + \frac{1}{2}mu^{2} = mgh_{f} +\frac{1}{2}mv^{2}$$

m is in every term. Let's divide by m.

$$gh_{i} + \frac{1}{2}u^{2} = gh_{f} +\frac{1}{2}v^{2}$$ $$(9.8)(30) + \frac{1}{2}(10)^{2} = (9.8)(0) + \frac{1}{2}v^{2}$$ $$294 + 50 = \frac{1}{2}v^{2}$$ $$344 = \frac{1}{2}v^{2}$$ $$\pm 26.2 \small\frac{m}{s} =\normalsize v$$
Example: A 0.20kg ball is placed on a spring compressed down 0.15m. The internet says that your spring has a spring constant of 200.0kg/s². How high could the ball go just using the energy stored in the spring?
solution $$U_{s} = U_{g}$$ $$\frac{1}{2}kx^{2} = mgh$$

Can't divide by mass this time.

$$\frac{1}{2}(200)(0.15)^{2} = (0.2)(9.8)h$$ $$1.15m = h$$
Example: A 0.20kg ball is placed on a spring compressed down 0.15m. The internet says that your spring has a spring constant of 200.0kg/s². How fast should the ball be moving right after it leaves the spring?
solution $$U_{s} = K$$ $$\frac{1}{2}kx^{2} = \frac{1}{2}mv^{2}$$

Can't divide by mass this time.

$$\frac{1}{2}(200)(0.15)^{2} = \frac{1}{2}(0.2)v^{2}$$ $$22.5 = v^{2}$$ $$\pm 4.74 \small\frac{m}{s} =\normalsize v$$

Energy Loss and Gain in a System


As the ball in the simulation falls, energy transfers from gravitational to kinetic. As it rises, energy transfers from kinetic to gravitational.

Energy is conserved. This means that if we add up the total energy of a system it will always be the same value, unless some energy enters or leaves the system we are tracking.

Turn friction on and observe what happens.


The kinetic energy gradually leaves the ball as it changes into a type of energy called heat.

Heat is the random motion of particles. More heat leads to higher temperatures.

So the internal energy of a system is always conserved, unless energy enters or leaves our system.

The first law of thermodynamics says that the change in internal energy of a system is equal to heat flow into a system plus work done on a system.

$$\Delta U = Q + W$$

\( U \) = total internal energy of a system [J]
\( Q \) = Heat, thermal energy added(+) or removed(-) from system [J]
\( W \) = work done on a system [J]
Example: Compressing a volume of gas does 100J of work, but at the same time the gas losses 200J of heat to the surroundings. What is the change in the internal energy of the gas? Will the gas get hotter or cooler in temperature?
solution $$\Delta U = Q + W$$ $$\Delta U = -200 + 100$$ $$\Delta U = -100J$$

The gas got cooler in temperature.

Example: A 0.43kg soccer ball kicked at 10m/s rolls down a 30m tall hill. We already solved this problem and got a final velocity of 26.2m/s for the ball.

When actually performing the experiment we found at the bottom of the hill the ball only has a speed of 20.0m/s. Calculate the energy loss for the ball. How could the ball have lost energy?
solution

Find the difference between the experimental final speed and the calculated final speed.

$$ \Delta v = 26.2-20.0$$ $$ \Delta v = 6.2$$

Convert the speed difference into kinetic energy.

$$KE = \frac{1}{2}mv^{2}$$ $$KE = \frac{1}{2}(0.43)(6.2)^{2}$$ $$KE = 8.26 J$$

The energy loss was probably from friction.

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