The correct answer is that the total energy at the top of the hill is equal to the total energy at the bottom of the hill. This is called the Law of Conservation of Energy. This law says that kinetic energy is equal to potential energy. We know that energy can not be created or destroyed, but that it can be transferred from one type to another. The total energy at the top of the hill is all potential energy. As the ball begins rolling down the hill, all the potential energy is converted to kinetic energy. Therefore, the total energy remains constant and energy is conserved.  

Determining expression for vertical component of ball's initial velocity:

\(\displaystyle sin(60^\circ)*v_i=v_y_i\)

\(\displaystyle \frac{\sqrt{3}}{2}*v_i=v_y_i\)

Using conservation of energy:

\(\displaystyle .5mv_y_i^2+mgh_i=.5mv_y_f^2+mgh_f\)

Solving for \(\displaystyle v_y_i\)

\(\displaystyle \sqrt{v_y_f^2+2gh_f-2gh_i}=v_y_i\)

At the maximum height, there will be no motion in the vertical direction, thus \(\displaystyle v_y_f=0\)

Plugging in values:

\(\displaystyle \sqrt{0^2+2*9.8*4-2*9.8*2}=v_y_i\)

\(\displaystyle v_y_i=6.26\frac{m}{s}\)

Using conservation of energy:

\(\displaystyle KE_1+EPE_1=KE_2+EPE_2\)

\(\displaystyle .5mv_1^2+.5kx_1^2=.5mv_2^2+.5kx^2\)

While at the maximum value of stretch (the amplitude), the velocity will be zero. While at the maximum velocity, the stretch will be zero.

Plugging in values:

\(\displaystyle .5*.075*2^2+.5k(.065)^2=.5*.075(0)^2+.5k*.065^2\)

Solving for \(\displaystyle k\)

\(\displaystyle k=\frac{71N}{m}\)

\(\displaystyle KE_1+PE_1=KE_2+PE_2\)

\(\displaystyle .5mv_i^2+mgh_i=.5mv_f^2+mgh_f\)

At the maximum height, velocity is zero.

\(\displaystyle .5mv_i^2+mgh_i=mgh_f\)

Solving for \(\displaystyle v_i\)

\(\displaystyle v_i=\sqrt{g(h_f-h_i)}\)

Plugging in values

\(\displaystyle v_i=\sqrt{9.8(3-1.5)}\)

\(\displaystyle v_i=3.83\frac{m}{s}\)

Based on the information given, his throw would have started at \(\displaystyle .5m\)above the ground. The pizza gained a maximum height of \(\displaystyle 3m\). Thus, the gravitational potential energy in relation to the starting position will be:

\(\displaystyle PE_{gravitational}=mg(h_f-h_i)\)

\(\displaystyle PE_{gravitational}=.55*9.8(3-.5)\)

\(\displaystyle PE_{gravitational}=13.5J\)

Since at it's maximum height, all of the energy will be gravitational, the work done on the pizza will be equal to the potential energy, thus:

\(\displaystyle W=13.5J\)

Conservation of energy:

\(\displaystyle .5mv_i^2+mgh_i=.5mv_f^2+mgh_f\)

Solving for \(\displaystyle v_i\)

\(\displaystyle \sqrt{v_f^2+2gh_f-2gh_i}=v_i\)

Plugging in values:

\(\displaystyle \sqrt{0^2+2*9.8*2-2*9.8*.6}=v_i\)

\(\displaystyle v_i=5.24\frac{m}{s}\)

This question tests your understanding of conservation of energy, gravitational potential energy, and energy of spring compression. In this question, a \(\displaystyle 4kg\) ball is dropped from an initial height of \(\displaystyle 20m\) from the ground, and then at the level of the ground, it contacts and compresses a spring, with all gravitational potential energy being transferred to the spring. You are then asked to calculate the spring constant. Because we are told that all gravitational potential energy is transferred to compressing the spring, we can set the gravitational potential energy equal to the spring compression energy, and solve for the spring constant as follows:

\(\displaystyle Energy_{total}= Energy_{initial}= Energy_{final}\)

\(\displaystyle \textup{Gravitational Potential Energy = Spring Compression Energy}}\)

\(\displaystyle \textup{mass * acceleration due to gravity x height = 0.5 * spring constant * distance of spring compression^{2}}\)

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

Next, we must isolate \(\displaystyle k\).

\(\displaystyle k= \frac{2mgh}{x^{2}}\)

\(\displaystyle k= \frac{(2 * 4 kg * 9.8 \frac{m}{s^{2}} * 20 m)} {(0.5 m)^{2}}\)

\(\displaystyle k= 6,272 \frac{N}{m}\)

Therefore, the spring constant is \(\displaystyle 6,272 \frac{N}{m}\).

This question tests your understanding of the concept of conservation of energy (energy can neither be created nor destroyed). In this example, initially all of the ball's energy is potential energy, as it is immobile at a height of \(\displaystyle 30m\) prior to being dropped. At the ball hits the ground, its energy is entirely transformed to kinetic energy. Thus, since we are assuming that no energy is lost to drag forces, the initial potential energy of the ball is equal to the final kinetic energy of the ball. We can solve for the ball's velocity as follows:

Potential Energy (initial) = Kinetic Energy (final)

\(\displaystyle \textup{mass}\ * \ \textup{acceleration due to gravity}\ *\ \textup{height} = \frac{1}{2}\ *\ \textup{mass}\ *\ \textup{final velocity}^2\)

\(\displaystyle mgh=\left(\frac{1}{2}\right)mv^2\)

Isolate the velocity term and solve.

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

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

\(\displaystyle v= \sqrt{\left(2 *9.8\frac{m}{s^2}* 30 m\right)}\)

\(\displaystyle v= \sqrt{588\frac{m^2}{s^s}}\)

\(\displaystyle v=24.2\frac{m}{s}\)

Thus, the speed of the ball as it hits the ground is \(\displaystyle 24.2\frac{m}{s}\)

This question tests your understanding of the concept of conservation of energy (energy can neither be created nor destroyed), and likewise of transitions between gravitational potential energy and kinetic energy. In this case, the \(\displaystyle 1kg\) toy car starts with a potential energy of \(\displaystyle 49 J\), since\(\displaystyle PE = mgh = 1kg * 9.8 \frac{m}{s^{2}} * 5m\). Thus, as this is the maximal height of the car, and no forces other than gravity are acting on the car, \(\displaystyle 49 J\) represents the maximal potential energy of the car.

We are then told that the car enters a loop with a radius of \(\displaystyle 2.5 m\), indicating that the maximal height of the loop is also \(\displaystyle 5.0 m\), and therefore, since no energy is lost to friction, the maximal potential energy of the car will still be \(\displaystyle 49 J\)at the peak of the ramp. We are explicitly asked to find the speed of the car when it is \(\displaystyle 1m\) off of the ground, on the loop.

There are a number of correct ways to calculate the speed of the car at this point. We will use the conservation of energy formula as follows:

\(\displaystyle Total Energy = Initial Gravitational Potential Energy\)

\(\displaystyle Initial Gravitational Potential Energy = PE (h = 1 m) + KE (h = 1 m)\)

\(\displaystyle E_{total} = 49 J = mgh_{h= 1 m} + \frac{1}{2}mv_{h=1m}^{2}\)

\(\displaystyle 49 J = (1 kg * 9.8 \frac{m}{s^{2}} * 1 m) + (\frac{1}{2} * 1 kg * v_{h=1m}^{2})\)

\(\displaystyle 49 J - 9.8 J = \frac{1}{2} kg *v_{h=1m}^{2}\)

\(\displaystyle 39.2J = \frac{1}{2} kg *v_{h=1m}^{2}\)

\(\displaystyle 78.4 \frac{J}{kg} = v_{h=1m}^{2}\)

\(\displaystyle v_{h=1m}= \sqrt{78.4 \frac{J}{kg}}\)

\(\displaystyle v_{h=1m}= 8.9 \frac{m}{s}\)

Thus, the speed of the car at a height of \(\displaystyle 1 m\) off of the ground on the loop is \(\displaystyle 8.9 \frac{m}{s}\).

This problem tests your knowledge of conservation of energy, gravitational force, and spring force. First, we begin with a \(\displaystyle 2kg\) block that is placed directly on a spring. We are then told that all of the gravitational energy on the block is transferred to the spring. Therefore, we can set the gravitational force on the block equal to the spring compression force as follows:

\(\displaystyle Force_{gravitational} = Force_{spring}\)

\(\displaystyle \textup{mass * acceleration due to gravity = - spring constant * compression distance}\)

\(\displaystyle mg=-kx\)

Next, we must isolate \(\displaystyle k\).

\(\displaystyle k= -\frac{mg}{x}\), and we can eliminate the negative sign since the question asks for the magnitude of the spring constant

\(\displaystyle k= \frac{(2 kg * 9.8 \frac{m}{s^{2}})} {0.2 m}\)

\(\displaystyle k = 98 \frac{N}{m}\)

Therefore, the spring constant is \(\displaystyle 98 \frac{N}{m}\).