Neglecting air resistance, we can say that the energy stored in the bungie when it is at its maximum stretched distance is equal to the initial gravitational potential energy of the jumper:

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

The initial height is the length of the bungie when it is fully stretched. We can write this out as:

\(\displaystyle h_i = l +x\)

\(\displaystyle l\) is the length of the unstretched bungie, and \(\displaystyle x\) is the distance the bungie stretches. Plugging this into the original expression, we get:

\(\displaystyle mg(l+x) = \frac{1}{2}kx^2\)

Rearranging for the mass of the jumper, we get:

\(\displaystyle m = \frac{kx^2}{2g(l+x)}\)

We have all of our values, allowing us to solve:

\(\displaystyle m = \frac{(100\frac{N}{m})(20m)^2}{2(10\frac{m}{s^2})(20m+20m)}\)

\(\displaystyle m = 50kg\)

First, we calculate the kinetic energy of the block as it slides:

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

\(\displaystyle KE=\frac{1}{2}(4kg)(2.5\frac{m}{s})^2=12.5J\)

We know that all of the kinetic energy is converted to spring potential energy during the collision. We can use the equation for the potential energy of a spring and set it equal to the kinetic energy. Then, solve for the compression distance:

\(\displaystyle KE=U_{spring}=\frac{1}{2}kx^2\)

\(\displaystyle 12.5J=\frac{1}{2}(30\frac{N}{m})(x)^2\)

\(\displaystyle x^2=\frac{12.5J}{\frac{1}{2}(30\frac{N}{m})}=0.833m^2\)

\(\displaystyle x=\sqrt{0.833m^2}=0.91m\)

This is a conservation of energy problem. First, we can set initial energy equal to final energy due to the law of conservation of energy. Therefore, \(\displaystyle E_{i}=E_{f}\) can be broken into its components:

\(\displaystyle PE_{i}+KE_{i}=PE_{f}+KE_{f}\)

\(\displaystyle mgh_{i}+\frac{1}{2}mv_{i}^{2}=mgh_{f}+\frac{1}{2}mv_{f}^{2}\)

Because the initial velocity is \(\displaystyle 0 \frac{m}{s}\) and the final height is 0 m, the equation can then be simplified:

\(\displaystyle mgh_{i}=\frac{1}{2}m{v_{f}}^{2} \rightarrow gh_{i}=\frac{1}{2}{v_{f}}^{2}\)

From here, we can rearrange the equation to solve for the final velocity:

\(\displaystyle gh_{i}=\frac{1}{2}{v_{f}}^{2}\)

\(\displaystyle {v_{f}}^{2}=2gh_{i}\)

\(\displaystyle v_{f}=\sqrt{2gh_{i}}\)

From here, we can plug in the known values and calculate the final velocity of the roller coaster car:

 \(\displaystyle v_{f}=\sqrt{2(9.81\frac{m}{s^2})(100m)}=44\frac{m}{s}\)

First, let's set up an axis for this problem. Have the positive direction be towards the outfield and the pitcher and the negative direction be towards home plate and the batter. Next, identify the given information: \(\displaystyle m=0.145kg\)

\(\displaystyle v_{i}=-31m/s\) (This value is negative because the ball is moving towards home plate.)

\(\displaystyle v_{f}=37m/s\) (This value is positive because the ball is moving towards the outfield.)

The impulse of the baseball is equal to its change in momentum:

\(\displaystyle I=\Delta p = p_{f}-p_{i}\)

Momentum is equal to mass times velocity, so this equation can be simplified further:

\(\displaystyle I=mv_{f}-mv_{i}=m(v_{f}-v_{i})\)

By plugging in the given information, we can solve for the impulse:

\(\displaystyle I=(0.145kg)(37\frac{m}{s}+31\frac{m}{s})=9.86N\cdot s\)

Use conservation of energy:

\(\displaystyle E_{intial}=E_{final}\)

\(\displaystyle PE_{spring-initial}+PE_{gravity-initial}=PE_{spring-final}+PE_{gravity-final}\)

Initially there is no gravitational energy, and in the final state there is no longer spring potential energy.

\(\displaystyle PE_{spring-initial}=PE_{gravity-final}\)

\(\displaystyle .5kx^2=mgh\)

Solve for the spring constant:

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

Plugging in values:

\(\displaystyle \frac{2*23*9.8*10}{.10^2}=k\)

\(\displaystyle 4.5*10^5\frac{N}{m}=k\)

The spring force is going to add to the gravitational force to equal zero.

\(\displaystyle mg+kx=0\)

Plugging in values:

\(\displaystyle 2.55*10^5*-9.8+k*.225=0\)

Solving for \(\displaystyle k\)

\(\displaystyle k=1.11*10^7\frac{N}{m}\)

Using

\(\displaystyle PE_{spring}=.5kx^2\)

Plugging in values:

\(\displaystyle PE_{spring}=1.11*10^7\frac{N}{m}*(.225m)^2\)

\(\displaystyle PE_{spring}=5.62*10^{5}\textup{ J}\)

Energy is conserved throughout this entire problem, before the ball is dropped all of the energy is in the form of potential energy represented by \(\displaystyle mgh\), halfway down there is potential and kinetic energy\(\displaystyle \left(\frac{1}{2}mv^{2}\right)\) present. Energy at the start must be equal to the energy halfway down which gives rise to the equation:

\(\displaystyle mgh_{_{0}}=mgh_{f}+\frac{1}{2}mv^{2}\)

where \(\displaystyle h_{0}\) is the initial height and \(\displaystyle h_{f}\) is the height halfway down. Mass can be cancelled out and the heights and gravity constant can be plugged in in order to find velocity. Plugging in the values:

\(\displaystyle 9.81*100=\frac{1}{2}v^{^{2}}+9.81(50)\)

\(\displaystyle v=31.3\frac{m}{s}\)

At the equilibrium position, all of the energy in the system is in the form of kinetic energy. When the spring is fully stretched, all of the system's energy is in spring potential energy. Since no energy is leaving the system, we can just set the kinetic energy equal to the potential to find the distance the spring will stretch. Kinetic energy is \(\displaystyle \frac{1}{2}mv^2\) and spring potential energy is \(\displaystyle \frac{1}{2}kx^2\) 

Setting these equal we get:

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

\(\displaystyle x=v\sqrt{\frac{m}{k}}\)

Plugging in our values we get:

\(\displaystyle (10\frac{m}{s})\sqrt{\frac{5kg}{50\frac{N}{m}}}=3.16m\)

Using conservation of energy:

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

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

Plugging in values:

\(\displaystyle .5.050*2^2+.5k(0)^2=.5*.050(0)^2+.5k*.05^2\)

Solving for \(\displaystyle k\)

\(\displaystyle k=\frac{80N}{m}\)