The running woman has kinetic energy as she is moving.

The candy bar has chemical potential energy.

The apple has gravitational potential energy.

The rubber band and the spring both have elastic potential energy.

Remember the law of conservation of energy: the total energy at the beginning equals the total energy at the end. In this case, we have only potential energy at the beginning and only kinetic energy at the end. (The initial velocity is zero, and the final height is zero).

\(\displaystyle PE_i=KE_f\)

If we can find the potential energy, we can find the kinetic energy. The formula for potential energy is \(\displaystyle PE=mgh\).

Using our given values for the mass, height, and gravity, we can solve using multiplication. Note that the height becomes negative because the book is traveling in the downward direction.

\(\displaystyle PE=2kg*-9.8\frac{m}{s^2}*-2.3m\)

\(\displaystyle PE=45.08J\)

The kinetic energy will also equal \(\displaystyle 45.08J\), due to conservation of energy.

Remember the law of conservation of energy: the total energy at the beginning equals the total energy at the end. In this case, we have only potential energy at the beginning and only kinetic energy at the end. (The initial velocity is zero, and the final height is zero).

\(\displaystyle PE_i=KE_f\)

If we can find the potential energy, we can find the kinetic energy. The formula for potential energy is \(\displaystyle PE=mgh\).

Using our given values for the mass, height, and gravity, we can solve using multiplication. Note that the height becomes negative because the book is traveling in the downward direction.

\(\displaystyle PE=1.12kg*-9.8\frac{m}{s^2}*-2.03m\)

\(\displaystyle PE=22.28J\)

The kinetic energy will also equal \(\displaystyle 22.28J\), due to conservation of energy.

For this problem, use the law of conservation of energy. This states that the total energy before the fall will equal the total energy after the fall. The initial kinetic energy will be zero, and the final potential energy will be zero; thus, the initial non-zero potential energy will be equal to the final non-zero kinetic energy.

\(\displaystyle PE_i+KE_i=PE_f+KE_f\)

\(\displaystyle PE_i=KE_f\)

From there, expand the equation to include the individual formulas for potential and kinetic energy calculation.

\(\displaystyle mgh_i=\frac{1}{2}mv_f^2\)

Notice that the mass will cancel out from both sides.

\(\displaystyle gh=\frac{1}{2}v_f^2\)

Now we can solve for the final velocity in terms of the initial height.

\(\displaystyle 2gh=v_f^2\)

\(\displaystyle \sqrt{2gh}=\sqrt{v_f^2}\)

\(\displaystyle \sqrt{2gh}=v_f\)

The potential energy at a given height it the product of the height, the mass of the object, and the acceleration of gravity.

Potential gravitational energy is given from the equation:

\(\displaystyle PE=mgh\)

The formula for potential energy is:

\(\displaystyle PE=mgh\)

Since \(\displaystyle g\) is a constant, the acceleration due to gravity on Earth, we only need the mass and the height. The problem gives a height, so we only need mass.

Given the mass and the height, we would be able to calculate the initial potential energy.

To answer this question, we can address each type of energy separately. There is no conservation of energy in this problem; kinetic energy is not converted to potential energy as the man runs up the hill. Instead, he is accelerating, indicating an outside force that disallows conservation of energy.

First, we will find the maximum potential energy using the equation:

\(\displaystyle PE = mgh\)

The man's mass and the acceleration of gravity will remain constant. The only changing variable is height. When the height is greatest, the potential energy will be the greatest. We can conclude that the potential energy will thus be greatest at the top of the hill.

Now we will look at the equation for kinetic energy:

\(\displaystyle KE = \frac{1}{2}mv^2\)

The man's mass will remain constant, and the only changing variable will be the velocity. We are told that the man accelerates as he runs up the hill, indicating that his velocity is increasing. This tells us that he will reach a maximum velocity when he reaches the top of the hill, at which point he maintains a steady velocity along the top of the hill. Since kinetic energy is at a maximum when velocity is at a maximum, we can conclude that kinetic energy is greatest at the top of the hill.

For this problem, we're going to use the law of conservation of energy. Since we're looking for max velocity, we're going to say that the \(\displaystyle PE=KE\) of the system.

The formula for potential energy of a spring is \(\displaystyle PE=\frac{1}{2}kx^2\)

Therefore:

\(\displaystyle PE=KE\)

\(\displaystyle \frac{1}{2}kx^2=\frac{1}{2}mv^2\)

Notice that the \(\displaystyle \frac{1}{2}\)'s cancel out.

Plug in our given values.

\(\displaystyle kx^2=mv^2\)

\(\displaystyle 180\frac{N}{m}*(0.89m)^2=2kg*v^2\)

\(\displaystyle 180\frac{N}{m}*0.7921m^2=2kg*v^2\)

\(\displaystyle 142.578J=2kg*v^2\)

\(\displaystyle \frac{142.578J}{2kg}=v^2\)

\(\displaystyle 71.289\frac{m^2}{s^2}=v^2\)

\(\displaystyle \sqrt{71.289\frac{m^2}{s^2}}=\sqrt{v^2}\)

\(\displaystyle 8.44\frac{m}{s}=v\)

Total mechanical energy is the sum of potential energy and mechanical energy.

\(\displaystyle ME=PE+KE\)

We can expand this equation to include the formulas for kinetic and potential energy.

\(\displaystyle ME=mgh+\frac{1}{2}mv_{iy}^2\)

Since we are only looking at vertical energies, we need to find the initial vertical velocity to apply toward the kinetic energy.

To find the vertical velocity we use the equation \(\displaystyle v_{i}* \sin(\Theta)=v_{iy}\).

We can plug in the given values for the angle and initial velocity to solve.

\(\displaystyle 30\frac{m}{s}* \sin(35^{\circ})=v_{iy}\)

\(\displaystyle 30\frac{m}{s}* (0.57)=v_{iy}\)

\(\displaystyle 17.21\frac{m}{s}=v_{iy}\)

Now we have all the terms necessary to solve for the total energy. Keep in mind that the change in height is going to be negative, since the rock is traveling downward.

\(\displaystyle ME=(0.5kg)(-9.8\frac{m}{s^2})(-10m)+\frac{1}{2}(0.5kg)(17.21\frac{m}{s})^2\)

\(\displaystyle ME=49J+74.05J\)

\(\displaystyle ME=123.05J\)

Mechanical energy is the sum of potential and kinetic energies.

\(\displaystyle ME=PE+KE\)

\(\displaystyle ME=mgh+\frac{1}{2}mv^2\)

Since we're only looking at the vertical components, we need to find the initial vertical velocity. This will be used for the vertical component of the kinetic energy. Use the sine function, initial velocity, and angle for this calculation.

\(\displaystyle v_{i}*\sin(\theta)=v_{iy}\)

\(\displaystyle 31\frac{m}{s}*\sin(20^{\circ})=v_{iy}\)

\(\displaystyle 31\frac{m}{s}*(0.34)=v_{iy}\)

\(\displaystyle 10.6\frac{m}{s}=v_{iy}\)

Gravitational potential energy only exists in the vertical plane, so we do not need to manipulate the values. Use our vertical velocity, mass, and the height of the building to find the vertical mechanical energy at the point of release.

\(\displaystyle ME=(0.5kg)(9.8\frac{m}{s^2})(12m)+\frac{1}{2}(0.5kg)(10.6\frac{m}{s})^2\)

\(\displaystyle ME=58.8J+28.09J\)

\(\displaystyle ME=86.89J\)