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:

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:

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

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

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:

Substituting in expressions for each force, we get:

Eliminating mass and rearranging for , we get:

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

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  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:

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:

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

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 .

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

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

The normal force is perpendicular to the plane:

First, we need to find .

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

Rearranging to solve this equation for , you get

Substituting in the side lengths of the given triangle, we can solve for .

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

because , or, in terms of this problem, .

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

The component of  perpendicular to the slope is , where  is the angle between the ground and the incline.

 where  is the normal force.

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

 is the coefficient of kinetic friction.

The component of  parallel to the slope is:

The net force on the block is:

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:

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:

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

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:

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:

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

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:

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:

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

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:

During terminal velocity:

Thus, by Newton's second law:

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 . Second, the normal force of the plane, which is given as . Third, the friction force, which acts opposite to its direction of motion and is given as . 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  is not along either of these axes, so we resolve the  force into its components,  along the negative y-axis, and  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: 

 and 

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 . 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 . Plugging these accelerations in, we find that  and 

Summing all the forces in the x-direction gives us 

Summing all the forces in the y-direction gives us

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

Solving for  in the second equation gives us . Thus the normal force is equal to the cosine component of the weight. Substituting  in for  in the first equation will give us the following:

Now we solve the equation for . Adding  to each side gives us: 

Now we divide each side by  to obtain:

The final result is obtained by canceling the  factor and using the triginometric identity:

Therefore we arrive at the conclusion that