Before we start using equations, we need to determine what forces are acting on the block in this system. The only relevant forces in this situation are gravity and friction. We are given the acceleration of the block, giving us the tools to find the net force.

Using Newton's second law, we can write:

\(\displaystyle F_{net}=ma\)

\(\displaystyle F_g -F_f = ma\)

The force of friction is subtracted because it is in the opposite direction of the movement of the block. Substituting in expressions for each variable, we get:

\(\displaystyle mgsin(\theta) - \mu_k mgcos(\theta) = ma\)

Canceling out mass and rearranging for the coefficient of kinetic friction, we get:

\(\displaystyle \mu_k = \frac{gsin(\theta)-a}{gcos(\theta)}\)

We have values for each variable, allowing us to solve:

\(\displaystyle \mu_k = \frac{(10\frac{m}{s^2})sin(60^{\circ})-7\frac{m}{s^2}}{(10\frac{m}{s^2})cos(60^{\circ})}\)

\(\displaystyle \mu_k = 0.33\)

There are two relevant forces acting on the block in this scenario: gravity and friction. We can use Newton's second law to solve this problem:

\(\displaystyle F_{net} = ma\)

\(\displaystyle F_g - F_f = ma\)

Substituting in expressions for each force, we get:

\(\displaystyle mgsin(\theta) - \mu_kmgcos(\theta) = ma\)

Eliminating mass and rearranging for \(\displaystyle g\), we get:

\(\displaystyle g = \frac{a}{sin(\theta)-\mu_kcos(\theta)}\)

At this point, we can plug in values for each variable and solve:

\(\displaystyle g = \frac{12\frac{m}{s^2}}{sin(40^{\circ})-(0.35)cos(40^{\circ})}\)

\(\displaystyle g = 18.4\frac{m}{s^2}\)

Start by drawing in the forces acting on each block. You could also draw in the force of gravity and the normal force for each block, but they have been omitted from the image because they cancel each other out for each block and because there is no friction in this problem.

We are given in the question that the force \(\displaystyle F\) is 100 N. Since the blocks are connected by a string, they will therefore accelerate at the same rate, and we can treat them as a system that moves as if it were one object of total mass 80 kg (30 kg plus 50 kg). Use Newton's second law:

\(\displaystyle F_{net}=ma\)

In this problem, the two tension forces form an action/reaction pair and therefore are equal in magnitude but opposite in direction (Newton's third law). So:

\(\displaystyle 100\:N-T+T=80\:kg\cdot a\)

We can solve for acceleration, since the tensions cancel out.

\(\displaystyle a=\frac{100\:N}{80\:kg}=1.25\:\frac{m}{s^2}\)

Now that we have acceleration, we need to write a new equation in which the tension force does not cancel out so that we can solve for the tension \(\displaystyle T\).

Do this by using Newton's second law again, except for only one of the blocks:

\(\displaystyle F_{net}=ma\)

Lets consider the 30 kg block. The only force acting on the 30 kg block is the tension \(\displaystyle T\), and the acceleration is what we found above.

\(\displaystyle T=30\:kg\cdot 1.25\:\frac{m}{s^2}=38\:N\)

The normal force is perpendicular to the plane:

First, we need to find \(\displaystyle \theta\).

We can solve for \(\displaystyle \theta\) using the trigonometric equation that applies in this instance. We know the length of the side opposite of \(\displaystyle \theta\) (5 m) and the length of the side adjacent to \(\displaystyle \theta\) (10 m), so we can use the following equation to solve for \(\displaystyle \theta\):

 \(\displaystyle tan\theta=\frac{l_{opposite}}{l_{adjacent}}\)

Rearranging to solve this equation for \(\displaystyle \theta\), you get

\(\displaystyle \theta=arctan(\frac{l_{adjacent}}{l_{opposite}})\)

Substituting in the side lengths of the given triangle, we can solve for \(\displaystyle \theta\).

\(\displaystyle \theta = \arctan\left(\frac{5}{10} \right)=26.6^\circ\)

Note that the normal force is one of the legs of another right triangle. The other leg is the parallel force, and the hypotenuse is the force of gravity.

Using trigonometry, we know that

\(\displaystyle F_n=F_g \cos{\theta}\)

because \(\displaystyle cos\theta=\frac{l_{adjacent}}{l_{hypotenuse}}\), or, in terms of this problem, \(\displaystyle l_{adjacent}=cos\theta\cdot l_{hypotenuse}\).

Substituting in the known values into this equation, we can solve for the normal force:

\(\displaystyle F_n=(9.81\frac{m}{s^2}\cdot m) \cos{26.6^\circ}=8.77\frac{m^2}{s^2} \cdot m\)

\(\displaystyle F=mg=5kg\left(10\frac{m}{s^2} \right)=50N\) 

The component of \(\displaystyle mg\) perpendicular to the slope is \(\displaystyle mg\cdot cos\theta\), where \(\displaystyle \theta\) is the angle between the ground and the incline.

\(\displaystyle 50N\cdot cos(30^{\text o}) = 43.3N = -F_N\) where \(\displaystyle F_N\) is the normal force.

The force of friction, which is in the direction opposing motion is:

\(\displaystyle F_{fr}=\mu_k F_{N}=0.2\cdot43.3N=8.66N\)

\(\displaystyle \mu_k\) is the coefficient of kinetic friction.

The component of \(\displaystyle mg\) parallel to the slope is:

\(\displaystyle mg\cdot sin\theta = 50N\sin 30^{\textup o} = 25N\)

The net force on the block is:

\(\displaystyle F_{net}=F_{perpendicular}-F_N+F_{parallel}-F_{fr}\)

\(\displaystyle F_{net}=43.3N-43.3N+25N-8.66N=16.34N\)

\(\displaystyle F_{net}=ma\)

\(\displaystyle 16.34N=5kg\cdot a\)

\(\displaystyle a=3.37\frac{m}{s^2}\) 

Begin by diagraming the forces acting on the mass in the problem:

The mass itself creates a force due to gravity in the downward direction:

\(\displaystyle \overrightarrow{W}=-mg\widehat{j}=-35kg(9.81\frac{m}{s^2})\widehat{j}\)

\(\displaystyle \overrightarrow{W}=-343.4N\widehat{j}\)

For the mass to remain stationary, the forces must be in equilibrium. Therefore the sum of forces in the x and y directions each must be zero:

\(\displaystyle F_y=T_{1y}+T_{2y}-W=0\)

\(\displaystyle F_x=T_{2x}-T_{1x}=0\)

The x and y tensions can be written in terms of the magnitude of the tension in each cable. Begin with the x direction:

\(\displaystyle T_2cos(30^{\circ})-T_1cos(30^{\circ})=0\)

\(\displaystyle T_1=T_2\)

Because the angle is the same for each, the tension in each angle must be equivalent. Use this property when performing the force balance in the y direction:

\(\displaystyle T_1sin(30^{\circ})+T_2sin(30^{\circ})-W=0\)

\(\displaystyle 2T_1sin(30^{\circ})=W\)

\(\displaystyle T_1=343.4N\)

Begin by drawing a force diagram of forces acting on the mass:

The mass itself creates a force due to gravity in the downward direction.

For the mass to remain stationary, the forces must be in equilibrium. Therefore the sum of forces in the x and y directions each must be zero:

\(\displaystyle F_y=T_{1y}+T_{2y}-W=0\)

\(\displaystyle F_x=T_{2x}-T_{1x}=0\)

The x and y tensions can be written in terms of the magnitude of the tension in each cable. Begin with the x direction:

\(\displaystyle T_2cos(60^{\circ})-T_1cos(60^{\circ})=0\)

\(\displaystyle T_1=T_2\)

Because the angle is the same for each, the tension in each angle must be equivalent. Use this property when performing the force balance in the y direction:

\(\displaystyle T_1sin(60^{\circ})+T_2sin(60^{\circ})-W=0\) 

\(\displaystyle 2T_2sin(60^{\circ})=W\)

\(\displaystyle 2T_2sin(60^{\circ})=9.81\frac{m}{s^2}(50kg)\)

\(\displaystyle T_2=\frac{1}{2sin(60^{\circ})}9.81\frac{m}{s^2}(50kg)\)

\(\displaystyle T_2=283.2N\)

Begin by diagraming the forces acting on the mass:

The mass itself creates a force due to gravity in the downward direction.

For the mass to remain stationary, the forces must be in equilibrium. Therefore the sum of forces in the x and y directions each must be zero:

\(\displaystyle F_y=T_{1y}+T_{2y}-W=0\)

\(\displaystyle F_x=T_{2x}-T_{1x}=0\)

The x and y tensions can be written in terms of the magnitude of the tension in each cable. Begin with the x direction:

\(\displaystyle T_2cos(30^{\circ})-T_1cos(60^{\circ})=0\)

\(\displaystyle T_2=\frac{1}{2cos(30^{\circ})}T_1\)

Because the angle is the same for each, the tension in each angle must be equivalent. Use this property when performing the force balance in the y direction:

\(\displaystyle T_1sin(60^{\circ})+T_2sin(30^{\circ})-W=0\) 

\(\displaystyle T_1sin(60^{\circ})+(\frac{1}{2cos(30^{\circ})}T_1)sin(30^{\circ})-W=0\)

\(\displaystyle T_1(sin(60^{\circ})+\frac{tan(30^{\circ})}{2})=W\)

\(\displaystyle T_1=\frac{1}{sin(60^{\circ})+\frac{tan(30^{\circ})}{2}}W\)

\(\displaystyle T_1=\frac{1}{sin(60^{\circ})+\frac{tan(30^{\circ})}{2}}(50kg)(9.81\frac{m}{s^2})\)

\(\displaystyle T_1=424.8N\)

During terminal velocity:

\(\displaystyle a=0\)

Thus, by Newton's second law:

\(\displaystyle F_{net}=ma=0\)

\(\displaystyle F_{net}=F_g+F_{air}=0\)

The free body diagram of the block is given above. This block has three forces acting on it. First, it's weight under the influence of gravity, which is given as \(\displaystyle mg\). Second, the normal force of the plane, which is given as \(\displaystyle F_N\). Third, the friction force, which acts opposite to its direction of motion and is given as \(\displaystyle \mu_kF_N\). We choose a coordinate system so that our x-axis aligns with the motion of the block down the plane, and the y-axis aligns with the direction of the normal force. Thus the friction force points in the negative direction of the x-axis, and the normal force is aligned with the positive direction of the y-axis. However, the weight \(\displaystyle mg\) is not along either of these axes, so we resolve the \(\displaystyle mg\) force into its components, \(\displaystyle mg\cos(\theta)\) along the negative y-axis, and \(\displaystyle mg\sin(\theta)\) along the positive x-axis. 

Now we can use Newton's 2nd law to relate the given forces above. Newton's 2nd law gives us two equations: 

\(\displaystyle \sum F_x=ma_x\) and \(\displaystyle \sum F_y=ma_y\)

Because the block is constrained to move along the surface of the inclined plane, there should be no acceleration in the y direction, and so \(\displaystyle a_y=0\). Also, because the block moves at constant velocity down the plane, Newton's 1st law assures us that there is no acceleration in the x direction as well, therefore \(\displaystyle a_x=0\). Plugging these accelerations in, we find that \(\displaystyle \sum F_x=0\) and \(\displaystyle \sum F_y=0\)

Summing all the forces in the x-direction gives us 

\(\displaystyle \sum F_x=mg\sin(\theta)-\mu_kF_N\) 

Summing all the forces in the y-direction gives us

 \(\displaystyle \sum F_y=F_N-mg\cos(\theta)\) 

Plugging these values into the force equations above gives us the following equations:

\(\displaystyle mg\sin(\theta)-\mu_kF_N=0\)

\(\displaystyle F_N-mg\cos(\theta)=0\)

Solving for \(\displaystyle F_N\) in the second equation gives us \(\displaystyle F_N=mg\cos(\theta)\). Thus the normal force is equal to the cosine component of the weight. Substituting \(\displaystyle mg\cos(\theta)\) in for \(\displaystyle F_N\) in the first equation will give us the following:

\(\displaystyle mg\sin(\theta)-\mu_kmg\cos(\theta)=0\)

Now we solve the equation for \(\displaystyle \mu_k\). Adding \(\displaystyle \mu_kmg\cos(\theta)\) to each side gives us: 

\(\displaystyle mg\sin(\theta)=\mu_kmg\cos(\theta)\)

Now we divide each side by \(\displaystyle mg\cos(\theta)\) to obtain:

\(\displaystyle \mu_k=\frac{mg\sin(\theta)}{mg\cos(\theta)}=\tan(\theta)\)

The final result is obtained by canceling the \(\displaystyle mg\) factor and using the triginometric identity:

\(\displaystyle \frac{\sin(\theta)}{\cos(\theta)}=\tan(\theta)\)

Therefore we arrive at the conclusion that \(\displaystyle \mu_k=\tan(\theta)\)