$\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), a normal force, and since it’s on an incline, a frictional}\\&\text{force to balance it all out:}\\&F_{mass}=m*g\\&F_{normal}=m*g*cos(a)\\&F_{friction}=m*g*sin(a)\\&F_{friction}=19kg(9.81\frac{m}{s^2})(sin(69^{\circ}))\\&F_{friction}=174.01N\end{align*}$

$\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), the force of the rope supporting it, a normal force,}\\&\text{and since it’s on an incline, a frictional force to balance}\\&\text{it all out. Notice that the tension in the rope mitigates some}\\&\text{of the gravitational force (as illustrated in the upper left}\\&\text{corner) to lessen what is applied to the slope:}\\&F_{mass}=m*g\\&F_{normal}=(F_{mass}-F_{rope})*cos(a)\\&F_{friction}=(F_{mass}-F_{rope})*sin(a)\\&F_{rope}=F_{mass}-\frac{F_{friction}}{sin(a)}\\&F_{rope}=2786.04N-\frac{207.04N}{sin(48^{\circ})}\\&F_{rope}=2507.44N\end{align*}$

$\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), the force of the rope supporting it, a normal force,}\\&\text{and since it’s on an incline, a frictional force to balance}\\&\text{it all out. Notice that the tension in the rope mitigates some}\\&\text{of the gravitational force (as illustrated in the upper left}\\&\text{corner) to lessen what is applied to the slope:}\\&F_{mass}=m*g\\&F_{normal}=(F_{mass}-F_{rope})*cos(a)\\&F_{friction}=(F_{mass}-F_{rope})*sin(a)\\&F_{friction}=(101kg(9.81\frac{m}{s^2})-693.57N)(sin(68^{\circ}))\\&F_{friction}=275.6N\end{align*}$

$\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), a normal force, and since it’s on an incline, a frictional}\\&\text{force to balance it all out:}\\&F_{mass}=m*g\\&F_{normal}=m*g*cos(a)\\&F_{friction}=m*g*sin(a)\\&F_{friction}=187kg(9.81\frac{m}{s^2})(sin(70^{\circ}))\\&F_{friction}=1723.84N\end{align*}$

$\displaystyle \begin{align*}&\text{To solve a problem of force balance, a good idea is to draw}\\&\text{it out. Since the block is at rest, i.e. stationary, the sum}\\&\text{of forces acting on it must be equal to zero. The forces acting}\\&\text{on it are gravity (in the form of the mass times the gravitational}\\&\text{constant), a normal force, and since it’s on an incline, a frictional}\\&\text{force to balance it all out:}\\&F_{mass}=m*g\\&F_{normal}=m*g*cos(a)\\&F_{friction}=m*g*sin(a)\\&F_{friction}=208kg(9.81\frac{m}{s^2})(sin(28^{\circ}))\\&F_{friction}=957.95N\end{align*}$

Set the gravitational potential energy and kinetic energy equal to each other and solve for the height.

$\displaystyle mgh = \frac{1}{2}mv^2$

Mass cancels.

$\displaystyle gh = \frac{1}{2}v^2$

Isolate the height and plug in our values.

$\displaystyle h = \frac{v^2}{2g} = \frac{(50\frac{m}{s})^2}{2*9.8\frac{m}{s^2}} = \frac{2500}{19.6} = 127.551m$

Rounding this gives $\displaystyle h = 128m$.

To find terminal velocity, set the magnitude of the drag force equal to the magnitude of the force of gravity since when these forces are equal and opposite, the object will stop accelerating:

$\displaystyle \small F_{d}=mg$

$\displaystyle \small bv^{2}=mg$

Solve for $\displaystyle v$

$\displaystyle v=\sqrt{\frac{mg}{b}}$

The ship's equation of velocity (found by solving the first-order differential equation) is $\displaystyle \small v=v_0e^{-kt}$

Substitute $\displaystyle t=\frac{2}{k}$ and solve.

Because the gravitational force depends only on the mass beneath the object (which is the gravitational version of Gauss's Law for charge), the acceleration steadily decreases as the object falls, and drops to zero at the center. Nevertheless, the velocity keeps increasing as the object falls, it just does so more slowly.

Relevant equations:

$\displaystyle E_{total}=K+U_g$

$\displaystyle K = \frac{1}{2}mv^2$

$\displaystyle U_g=\frac{-GMm}{r}$

For the rocket to escape the moon's gravity, its minimum total energy is zero. If the total energy is zero, the rocket will have zero final velocity when it is infinitely far from the moon. If total energy is less than zero, the rocket will fall back to the moon's surface. If total energy is greater than zero, the rocket will have some final velocity when it is infinitely far away. 

For the minimum energy case as the rocket leaves the surface:

$\displaystyle 0 = K + U_g = \frac{1}{2}mv_{esc}^2 - \frac{GMm}{r}$

Rearrange energy equation to isolate the velocity term.

$\displaystyle v_{esc}^2 = \frac{GMm}{\frac{1}{2}mr}$

$\displaystyle v_{esc}=\sqrt{\frac{2GM}{r}}$

Substitute in the given values to solve for the velocity.

$\displaystyle v_{esc}=\sqrt{\frac{2(6.67*10^{-11}\frac{m^2N}{kg^2})(7.2*10^{22}kg)}{(1.8*10^6m)}}=2310\frac{m}{s}\approx 2.3\frac{km}{s}$