A horizontal force, F1 = 85 N, and a force, F2 = 18.4 N acting at an angle of θ to the horizontal, are applied to a block of mass m = 2.1 kg. The coefficient of kinetic friction between the block and the surface is μk = 0.2. The block is moving to the right.

A. Solve numerically for the magnitude of the normal force, FN in Newtons, that acts on the block if θ = 30°.

B. Solve numerically for the magnitude of acceleration of the block, a in m/s2, if θ = 30°.

A. Well, I'm no mathematician, but let's give it a shot. To find the normal force, we need to consider the forces acting in the vertical direction. In this case, it's just the weight of the block and the vertical component of force F2. The weight of the block is given by mg, where m is the mass and g is the acceleration due to gravity.

So, we have Fn - m*g*cos(theta) = 0

Solving for Fn, we get Fn = m*g*cos(theta).

Plugging in the values, m = 2.1 kg, g ≈ 9.8 m/s^2, and theta = 30°, we can calculate Fn.

Fn = 2.1 kg * 9.8 m/s^2 * cos(30°)

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B. Now, let's find the magnitude of acceleration of the block. To do that, we need to consider the forces acting in the horizontal direction. We have the force F1 and the horizontal component of force F2, as well as the force of kinetic friction.

The force of kinetic friction is given by fk = μk*Fn, where μk is the coefficient of kinetic friction.

So, we have F1 + F2*cos(theta) - fk - m*a = 0

But we already know that fk = μk*Fn, so we can substitute that into the equation:

F1 + F2*cos(theta) - μk*Fn - m*a = 0

Now we can solve for the magnitude of acceleration, a:

a = (F1 + F2*cos(theta) - μk*Fn)/m

Plugging in the values, F1 = 85 N, F2 = 18.4 N, μk = 0.2, m = 2.1 kg, and theta = 30°, you can figure out the magnitude of acceleration, a.

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To solve for the magnitude of the normal force, FN, we need to consider the forces acting on the block in the vertical direction. Since the block is moving horizontally to the right, there is no net force acting in the vertical direction.

The vertical forces acting on the block are the weight of the block (mg) and the vertical component of the applied force (F1sinθ). The normal force (FN) provides an equal and opposite force to balance the weight and vertical component of the applied force.

Using the equilibrium equation in the vertical direction:

FN + F1sinθ - mg = 0

Given:
F1 = 85 N
θ = 30°
m = 2.1 kg
g = 9.8 m/s^2

We can plug in the known values and solve for FN:

FN + 85sin(30°) - 2.1 * 9.8 = 0
FN + 85 * 0.5 - 2.1 * 9.8 = 0
FN + 42.5 - 20.58 = 0
FN = 20.58 - 42.5
FN = -21.92 N

Since the negative value doesn't make physical sense for a normal force, we can disregard it. Therefore, the magnitude of the normal force is approximately 21.92 N.

To solve for the magnitude of the acceleration, a, we need to consider the forces acting on the block in the horizontal direction. The horizontal forces acting on the block are the applied force, F1cosθ, and the force of kinetic friction, fk.

Using Newton's second law in the horizontal direction:

F1cosθ - fk = ma

The force of kinetic friction, fk, can be calculated using the equation fk = μk * FN, where μk is the coefficient of kinetic friction.

Given:
F1 = 85 N
θ = 30°
m = 2.1 kg
μk = 0.2
FN = 21.92 N (from previous calculation)

We can plug in the known values and solve for a:

F1cosθ - μk * FN = ma
85cos(30°) - 0.2 * 21.92 = 2.1a
85 * 0.866 - 0.2 * 21.92 = 2.1a
73.61 - 4.384 = 2.1a
69.226 = 2.1a
a ≈ 32.94 m/s^2

Therefore, the magnitude of the acceleration of the block is approximately 32.94 m/s^2.

To solve for the magnitude of the normal force (FN) and the magnitude of acceleration (a), we can apply Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (Fnet = ma).

Let's start with finding the normal force (FN) using the given information and the angle θ = 30°.

A. Solving for the magnitude of the normal force (FN):

1. Determine the vertical component of the applied force (F2):
F2_vertical = F2 * sin(θ)
= 18.4 N * sin(30°)
≈ 9.2 N

2. Calculate the weight of the block (mg):
Weight = m * g
= 2.1 kg * 9.8 m/s^2 (assuming g is the acceleration due to gravity)
≈ 20.58 N

3. Determine the vertical component of the weight:
Weight_vertical = Weight * cos(θ)
= 20.58 N * cos(30°)
≈ 17.838 N

4. Calculate the normal force (FN):
FN = Weight_vertical + F2_vertical
= 17.838 N + 9.2 N
≈ 27.038 N

Therefore, the magnitude of the normal force (FN) that acts on the block is approximately 27.038 N when θ = 30°.

B. Solving for the magnitude of the acceleration (a):

1. Determine the horizontal component of the applied force (F2):
F2_horizontal = F2 * cos(θ)
= 18.4 N * cos(30°)
≈ 15.971 N

2. Calculate the frictional force (Ffriction):
Ffriction = μk * FN
= 0.2 * 27.038 N
≈ 5.408 N

3. Determine the net horizontal force:
Fnet_horizontal = F1 - F2_horizontal - Ffriction
= 85 N - 15.971 N - 5.408 N
≈ 63.621 N

4. Apply Newton's second law to find the acceleration (a):
Fnet = m * a
63.621 N = 2.1 kg * a
a ≈ 30.296 m/s^2

Therefore, the magnitude of the acceleration (a) of the block is approximately 30.296 m/s^2 when θ = 30°.

assuming the force is acting downward?

frictionforce=forcenormal*mu
= (F1*g + F2*sinTheta)mu

acceleration
net force=ma
F2cosTheta-forcefriction=ma
solve for a.