We can use conservation of energy to solve this problem.  Let us consider the types of energy at the beginning and the end.  At the beginning the child has gravitational potential energy at the top of the slide.  When the child reaches the bottom, the child has both kinetic energy and thermal energy as some energy was converted to heat because of the friction on the slide.

The law of conservation of energy states that we can set the energy at the beginning equal to the energy at the end.

\(\displaystyle GPE = KE + E_{thermal}\)

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

\(\displaystyle (22.3kg)(9.8m/s^2)(3.6m) = \frac{1}{2}(22.3kg)(2.1m/s)^2 + E_{thermal}\)

\(\displaystyle 786.74J = 49.17J + E_{thermal}\)

\(\displaystyle 737.57J = E_{thermal}\)

This \(\displaystyle 737.57J\) difference between the \(\displaystyle GPE\) at the top and the \(\displaystyle KE\) at the bottom is the energy lost to friction.

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.

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

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

\(\displaystyle 2gh = v_f\)

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.

We can use conservation of energy to analyze this problem.  When both balls are thrown off the building, they both have the same initial gravitational potential energy.  Additionally, since both balls are thrown with the same magnitude of the speed, they both have the same kinetic energy.  Energy is a scalar quantity and therefore does not have a direction.  Therefore it is independent of the path taken.

At the bottom, both balls will have converted all their potential energy and kinetic energy that they started with to kinetic energy at the bottom of the building.  Since both started with the same amount of total energy at the beginning, their kinetic energy at the end will also be the same.  Since both balls are assumed to have the same mass, their magnitude of their velocity (speed) will be the same as well.  However, their velocity will be in different directions because of the way that the ball was launched.

Knowns:

\(\displaystyle h = 250m\)

\(\displaystyle V_0 = 100m/s\)

Unknowns:

\(\displaystyle V_f = ??\)

The easiest way to solve this problem is by analyzing the energy of the object at different points during its path.  The first point to analyze is the moment it is launched from the top of the cliff.  Assuming that the ground is our zero point in reference, the projectile has both Gravitational Potential  Energy and Kinetic Energy when it is launched.

\(\displaystyle Total \, energy \, at \, launch = GPE + KE\)

At the end when the projectile lands on the ground, it no longer has gravitational potential energy.  It now is all \(\displaystyle KE\)

\(\displaystyle Total \, energy \, when \, it \, hits\, the \, ground = KE\)

We know that based on the law of conservation of energy, the total energy at the beginning must equal the total energy at the end.  Therefore

\(\displaystyle GPE_0 +KE_0 = KE_f\)

Our equations for each are as follows

\(\displaystyle GPE = mgh\)

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

We can substitute these equations in.

\(\displaystyle mgh + \frac{1}{2} mv_0^2 = \frac{1}{2} mv_f^2\)

Notice that mass is in every term so we can cancel it out.  That is why mass does not matter for this problem.

\(\displaystyle gh + \frac{1}{2} v_0^2 = \frac{1}{2} v_f^2\)

Substitute in our values and solve for the final velocity.

\(\displaystyle (9.8m/s^2)(250m) + \frac{1}{2} (100m/s)^2 = \frac{1}{2} v_f^2\)

\(\displaystyle v_f^2 =14900\)

\(\displaystyle v_f= 122 m/s\)

First, we need to determine how fast the roller coaster must be going at the top of the loop to continue in a circular motion.  At the top of the loop, the only force acting on the car is gravity.  Therefore the gravitational force must be the cause of the centripetal motion.

\(\displaystyle F_c = F_g\)

We know that the force of gravity is

\(\displaystyle F_g = mg\) 

And the centripetal force equation is

\(\displaystyle F_c = \frac{mv^2}{r}\)

We can set these two equations equal to each other.

\(\displaystyle mg = \frac{mv^2}{r}\)

Since mass is on both sides of the equation we can cancel it out.

\(\displaystyle g = \frac{v^2}{r}\)

We can rearrange and solve this equation for the velocity.

\(\displaystyle gr = v^2\)

\(\displaystyle \sqrt{gr} = v\)

\(\displaystyle \sqrt{(9.8m/s^2)(15m)} = v\)

\(\displaystyle 12.12m/s = v\)

We can now use the conservation of energy to determine the initial height of the roller coaster.  We know at the top of the roller coaster, there is only \(\displaystyle GPE\).  At the top of the loop of the coaster there is both \(\displaystyle KE\) and \(\displaystyle GPE\).

\(\displaystyle GPE_{top} = GPE_{loop} + KE_{loop}\)

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

Since mass is each factor, we can cancel it out.

\(\displaystyle gh = gd + \frac{1}{2}v^2\)

\(\displaystyle (9.8m/s^2)h = (9.8m/s^2)(30m) + \frac{1}{2}(12.12m/s)^2\)

\(\displaystyle (9.8m/s^2)h = 294 + 73.5\)

\(\displaystyle (9.8m/s^2)h = 367.5\)

\(\displaystyle h = 37.5m\)

The height of the coaster must start at \(\displaystyle 37.0m\).

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.

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

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

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