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# Conservation of Energy

## The amount of energy in a closed system never changes.

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Practice Conservation of Energy

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Conservation of Energy

Students will learn how to apply energy conservation in a closed system.

### Key Equations

\begin{align*} \sum E_{\text{initial}} = \sum E_\text{final} \; \; \text{The total energy does not change in closed systems}\end{align*}

Guidance
Energy is conserved in a closed system. That is, if you add up all the energy of an object(s) at one time it will equal all the energy of said object(s) at a later time. A closed system is a system where no energy is transferred in or out. The total energy of the universe is a constant (i.e. it does not change). The problems below do not consider the situation of energy transfer (called work). So friction and other sources where energy leaves the system are not present. Thus, one simply adds up all the potential energy and kinetic energy before and sets it equal to the addition of the total potential energy and kinetic energy after.

#### Example 1

Billy is standing at the bottom of a ramp inclined at 30 degrees. Billy slides a 2 kg puck up the ramp with an initial velocity of 4 m/s. How far up the ramp does the ball travel before it begins to roll back down? Ignore the effects of friction.

##### Solution

The potential energy of the puck when it stops at the top of it's path will be equal to the kinetic energy that it was initially rolled with. We can use this to determine the how high above the ground the puck will be above the ground when it stops, and then use trigonometry to find out how far up the ramp the puck will be when it stops.

\begin{align*} PE_i + KE_i &= PE_f + KE_f && \text{start with conservation of energy}\\ 0 + \frac{1}{2}mv^2 &= mgh + 0 && \text{take out the energy terms we know will be zero and substitute the equations for potential and kinetic energy.}\\ \frac{1}{2}v^2 &= gh && \text{simplify the equation}\\ h&=\frac{v^2}{2g} && \text{solve for h}\\ h&=\frac{(4\;\text{m/s})^2}{2*9.8\;\text{m/s}^2} && \text{substitute in the known values}\\ h&=0.82\;\text{m}\\ \end{align*}

Now we can find the distance up the ramp the ball traveled since we know the angle of the ramp and the height of the ball above the ground.

\begin{align*} \sin(30)&=\frac{h}{x}\\ x&=\frac{h}{\sin(30)}\\ x&=\frac{0.82\;\text{m}}{\sin(30)}\\ x&=1.6\;\text{m}\\ \end{align*}

### Time for Practice

1. A stationary bomb explodes into hundreds of pieces. Which of the following statements best describes the situation?
1. The kinetic energy of the bomb was converted into heat.
2. The chemical potential energy stored in the bomb was converted into heat and gravitational potential energy.
3. The chemical potential energy stored in the bomb was converted into heat and kinetic energy.
4. The chemical potential energy stored in the bomb was converted into heat, sound, kinetic energy, and gravitational potential energy.
5. The kinetic and chemical potential energy stored in the bomb was converted into heat, sound, kinetic energy, and gravitational potential energy.
2. A 1200 kg> car traveling with a speed of 29 m/s drives horizontally off of a 90 m cliff.
1. Sketch the situation.
2. Calculate the potential energy, the kinetic energy, and the total energy of the car as it leaves the cliff.
3. Make a graph displaying the kinetic, gravitational potential, and total energy of the car at each 10 m increment of height as it drops
3. A roller coaster begins at rest \begin{align*}120 \;\mathrm{m}\end{align*} above the ground, as shown. Assume no friction from the wheels and air, and that no energy is lost to heat, sound, and so on. The radius of the loop is \begin{align*}40 \;\mathrm{m}\end{align*}.
1. Find the speed of the roller coaster at points \begin{align*}B, C, D, E, F\end{align*}, and \begin{align*}H\end{align*}.
2. Assume that \begin{align*}25\end{align*}% of the initial potential energy of the coaster is lost due to heat, sound, and air resistance along its route. How far short of point \begin{align*}H\end{align*} will the coaster stop?
4. A pendulum has a string with length 1.2 m. You hold it at an angle of 22 degrees to the vertical and release it. The pendulum bob has a mass of 2.0 kg.
1. What is the potential energy of the bob before it is released? (Hint: use geometry to determine the height when released.)
2. What is its speed when it passes through the midpoint of its swing?
3. Now the pendulum is transported to Mars, where the acceleration of gravity g is \begin{align*}2.3 \;\mathrm{m/s}^2\end{align*}. Answer parts (a) and (b) again, but this time using the acceleration on Mars.

1. d
2. b. \begin{align*}KE = 504,600 \;\mathrm{J}; U_g = 1,058,400 \;\mathrm{J}; E_{total} = 1,563,000 \;\mathrm{J}\end{align*}
3. a. \begin{align*}34 \;\mathrm{m/s \ at \ B}; 49 \;\mathrm{m/s \ at \ C \ and \ F}; 28 \;\mathrm{m/s \ at \ D}, 40 \;\mathrm{m/s \ at \ E}, 0 \;\mathrm{m/s \ at \ H}\end{align*} b. it will make it up to only 90m, so 30m short of point H
4. a. 1.7 J b. 1.3 m/s c. 0.4 J, 0.63 m/s

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