Force is given by:

$\displaystyle F_{net}=m*a$, where $\displaystyle m$ is mass and $\displaystyle a$ is acceleration. 

We're given mass, but we aren't given acceleration. Since the question asks for average force $\displaystyle F_{average}$, we can determine average acceleration $\displaystyle a_{aver}$. 

$\displaystyle a_{aver}=\frac{\Delta v}{\Delta t}$, where $\displaystyle \Delta v$ is the change in velocity and $\displaystyle \Delta t$ is change in time. 

In our problem, $\displaystyle \Delta v= 3\frac{m}{s}$ and $\displaystyle \Delta t= 3s$

$\displaystyle a_{aver}=\frac{\Delta v}{\Delta t}=\frac{3\frac{m}{s}}{3s}=1\frac{m}{s^2}$

$\displaystyle F_{average}=m*a_{average}=5kg*1\frac{m}{s^2}=5N$

Here we must use the following formula:

$\displaystyle F=ma$

We can substitute our known values of mass and force and solve for acceleration

$\displaystyle a = \frac{F}{m}= \frac{8N}{4kg}= 2\frac{m}{s^2}$

Since we know the acceleration and the time it acts upon the object, we can determine the final velocity through the following equation:

$\displaystyle v=at=(2\frac{m}{s^2})(3s)= 6\frac{m}{s}$

If the block is to remain at rest on the plane, we know that the sum of the forces acting a long the plane must be equal and opposite. This means that the gravitational force acting along the plane is equal to and opposite of the force of friction. This can be demonstrated as:

$\displaystyle mg\sin{\theta}=\mu_sN=\mu_smg\cos{\theta}$

This can be rewritten as:

$\displaystyle \tan{\theta}=\mu_s$

For the block to remain at rest, the force of static friction must exceed (for this problem we will set them equal to each other since it gives us the best approximation); solve for the angle:

$\displaystyle \tan{\theta}=\mu_s$

$\displaystyle \theta=\tan{^{-1}\mu_s}$

$\displaystyle \theta=\tan{^{-1}\(0.4)}=21.8^{\circ}$

Newton's second law states:

$\displaystyle F=ma$

Where $\displaystyle F$ is the net force exerted upon an object, $\displaystyle m$ is the mass of the object and $\displaystyle a$ is the acceleration of the object.

We rearrange this equation to show:

$\displaystyle \frac{F}{m}=a$

Plug in our given values with $\displaystyle F=15N$ and $\displaystyle m=5kg$:

$\displaystyle \frac{15N}{5kg}=3\frac{m}{s^2}$

By Newton's second law:

$\displaystyle F=m*a$, where $\displaystyle F$ is force, $\displaystyle m$ is mass, and $\displaystyle a$ is acceleration.

$\displaystyle 20N= 10kg*a$

$\displaystyle a= 2\frac{m}{s^2}$ 

This is the acceleration. Since we're assuming this acceleration is constant over time, we can model velocity $\displaystyle v$ as: 

$\displaystyle v=a*t+v_0$ where $\displaystyle v_0$ is the initial velocity. 

Since the initial velocity is $\displaystyle 0$ in our problem, 

$\displaystyle v=2*t$ 

After $\displaystyle 5$ seconds, 

$\displaystyle v=2\frac{m}{s^2}*5s=10\frac{m}{s}$

Use work:

$\displaystyle W=F*d=E_i-E_f$

All energy will be kinetic.

$\displaystyle F*d=KE_i-KE_f$

$\displaystyle F*d=.5mv_f^2-.5mv_i^2$

Convert $\displaystyle \frac{km}{hr}$ to $\displaystyle \frac{m}{s}$:

$\displaystyle \frac{50km}{hr}*\frac{1000m}{1km}*\frac{1 hr}{3600 s}=13.9\frac{km}{hr}$

$\displaystyle \frac{0km}{hr}*\frac{1000m}{1km}*\frac{1 hr}{3600 s}=0\frac{km}{hr}$

Plug in values. Force will be negative as it is pointing against the direction of travel:

$\displaystyle F*1000=.5*5.11*10^{11}*0^2-.5*5.11*10^{11}*(13.9)^2$

Solve for $\displaystyle F$:

$\displaystyle F=-4.93*10^{10}$

$\displaystyle |F|=4.93*10^{10}$

Use Newton's second law:

$\displaystyle F=ma$

Plug in values:

$\displaystyle 5.11*10^{11}*a=4.93*10^{10}$

Solve for $\displaystyle a$:

$\displaystyle a=\frac{.096m}{s^2}$

Convert to $\displaystyle g$

$\displaystyle \frac{.096m}{s^2}*\frac{g}{\frac{9.8m}{s^2}}=.0098g$

Use work:

$\displaystyle W=F*d=E_i-E_f$

All energy will be kinetic.

$\displaystyle F*d=KE_i-KE_f$

$\displaystyle F*d=.5mv_f^2-.5mv_i^2$

Convert $\displaystyle \frac{km}{hr}$ to $\displaystyle \frac{m}{s}$:

$\displaystyle \frac{50km}{hr}*\frac{1000m}{1km}*\frac{1 hr}{3600 s}=13.9\frac{km}{hr}$

$\displaystyle \frac{0km}{hr}*\frac{1000m}{1km}*\frac{1 hr}{3600 s}=0\frac{km}{hr}$

Plug in values. Force will be negative as it is pointing against the direction of travel:

$\displaystyle F*1000=.5*5.11*10^{11}*0^2-.5*5.11*10^{11}*(13.9)^2$

Solve for $\displaystyle F$:

$\displaystyle F=-4.93*10^{10}$

$\displaystyle |F|=4.93*10^{10}$

Use frictional force:

$\displaystyle F_{fric}=\mu*m*g$

Plug in values:

$\displaystyle 4.93*10^{10}=\mu*5.11*10^{11}*9.8$

Solve for $\displaystyle \mu$

$\displaystyle .0098=\mu$

In this question, we're told that a horizontal force is being applied to an object on the moon. We're given the magnitude of this force and the mass of the object.

We're also given the acceleration due to gravity on the moon, but this is useless information for the purposes of this question. Since the force under consideration is acting horizontally on the object, and the force due to gravity acts vertically, these two forces are treated separately; the vertical acting force of gravity will not affect the horizontal kinematics of the object's movement.

Thus, this reduces to a simple application of Newton's second law, and we can use the following equation:

$\displaystyle F=ma$

Rearranging, we can isolate the term for acceleration:

$\displaystyle a=\frac{F}{m}$

Then plug in the values that we have to solve for our answer:

$\displaystyle a=\frac{180\: N}{20\: kg}=9\: \frac{m}{s^{2}}$

This problem can be solved using the expression relating torque and angular acceleration:

$\displaystyle \tau_{net} = I_{net}\alpha$

We can determine the total net torque on the system using the Newton's second law.

$\displaystyle F_{net}=ma=mg$

Applying this to both masses and assuming a force in the downward direction is positive, we get:

$\displaystyle F_{A}= m_Ag = (20kg)(10\frac{m}{s^2})=200N$

$\displaystyle F_B = m_Bg=(15kg)(10\frac{m}{s^2})=150N$

Then using the expression for torque, we get:

$\displaystyle \tau=Fd$

Where d is half the length of the rod.

$\displaystyle \tau = F\left ( \frac{1}{2}l\right )$

Applying this to both masses and assuming torque in the counterclockwise direction is positive, we get:

$\displaystyle \tau_A = F_A\left ( \frac{1}{2}l\right )$

$\displaystyle \tau_A = \left (200N \right )\left ( \frac{1}{2}(6m)\right )=600N\cdot m$

$\displaystyle \tau_B = -F_B\left ( \frac{1}{2}l\right )$

$\displaystyle \tau_B = \left (150N \right )\left ( \frac{1}{2}(6m)\right )=-225N\cdot m$

$\displaystyle \tau_{net}= \tau_A+\tau_B = 600N\cdot m -225N\cdot m = 375N\cdot m$

Now let's go back to the original equation:

$\displaystyle \tau_{net}=I_{net}\alpha$

Now we need to calculate the net moment of inertia:

 $\displaystyle I=mr^2= m_Ar^2+m_Br^2 = (m_A+m_B)r^2$

Where r is half the length of the rod:

$\displaystyle I= (m_A+m_B)\left ( \frac{1}{2}l\right )^2$

$\displaystyle I = (20kg + 15kg)\left ( \frac{1}{2}(6m)\right )^2$

$\displaystyle I = 315kg\cdot m^2$

Going back to the original equation and rearranging for angular acceleration:

$\displaystyle \alpha=\frac{\tau_{net}}{I_{net}} = \frac{375N\cdot m}{315kg\cdot m^2} = 1.19\frac{1}{s^2}$

Then relating this to linear acceleration using:

$\displaystyle a = r\alpha$

Where r is half the length of the rod again:

$\displaystyle a = \left ( \frac{1}{2}l\right )\alpha$

We have all of our values, so we can calculate the final answer:

$\displaystyle a = \left ( \frac{1}{2}(6m)\right )\left ( 1.19\frac{1}{s^2}\right )$

$\displaystyle a = 3.6\frac{m}{s^2}$

We can use Newton's second law:

$\displaystyle F_{net}=ma$

There are two forces acting on the system: gravity on each mass. For simplicity sake, we will examine these forces with respect to mass. Therefore, the force on mass B becomes an upward tension force on mass A. Now we need to clarify which direction is positive. For this problem, we'll say that a downward force is positive, and an upward force is negative.

We'll start with the gravitational force:

$\displaystyle F_g = m_Ag$

Now for tension. The tension force is simply the gravitational force applied to mass B:

$\displaystyle F_T = -m_Bg$

Note that it's negative because it is in the upward direction

Now adding these together to get the net force:

$\displaystyle F_{net}= m_Ag-m_Bg = g(m_A-m_B)$

Substituting this back into the original equation, we get:

$\displaystyle ma = g(m_A-m_B)$

Where m is the combination of both masses:

$\displaystyle (m_A+m_B)a=(m_A-m_B)g$

Rearrange for acceleration:

$\displaystyle a = \frac{g(m_A -m_B)}{m_A+m_B}$

We know each value, so we can solve the problem

$\displaystyle a = \frac{\left ( 10\frac{m}{s^2}\right )(20kg-12kg)}{20kg+12kg}$

$\displaystyle a = 2.5\frac{m}{s^2}$