Let's analyze and solve the problem step by step. The problem involves calculating:

1. The magnitude of the resistive frictional force.
2. The coefficient of kinetic friction ($\mu_k$) when the pulling angle $\alpha = 30^\circ$.

Step-by-step solution:

1. Understand the given information:

• Force applied ($F$): 90.0 N
• Mass of the object ($m$): 25.0 kg
• Angle of pulling ($\alpha$): $30^\circ$
• Condition: The object moves with constant velocity ($a = 0$).

Since the object is moving at constant velocity, the net force in the horizontal direction is zero ($F_{net} = 0$). This means the pulling force equals the frictional force.

2. Resolve the pulling force into components:

• The horizontal component of the pulling force: $F_x = F \cdot \cos(\alpha)$
• The vertical component of the pulling force: $F_y = F \cdot \sin(\alpha)$

3. Normal force ($N$):

The normal force is reduced by the vertical component of the pulling force. It is calculated as: $N = m \cdot g - F_y$ where $g$ is the acceleration due to gravity ($g = 9.81 \, \text{m/s}^2$).

4. Frictional force ($F_{\text{friction}}$):

The object is moving at constant velocity, so: $F_{\text{friction}} = F_x$

5. Coefficient of kinetic friction ($\mu_k$):

Using the frictional force formula: $F_{\text{friction}} = \mu_k \cdot N$ Rearranging for $\mu_k$: $\mu_k = \frac{F_{\text{friction}}}{N}$

Substituting the values:

1. Calculate $F_x$ and $F_y$: $F_x = 90.0 \cdot \cos(30^\circ) = 90.0 \cdot 0.866 = 77.94 \, \text{N}$ $F_y = 90.0 \cdot \sin(30^\circ) = 90.0 \cdot 0.5 = 45.0 \, \text{N}$

2. Calculate the normal force: $N = m \cdot g - F_y = (25.0 \cdot 9.81) - 45.0 = 245.25 - 45.0 = 200.25 \, \text{N}$

3. The frictional force is equal to $F_x$: $F_{\text{friction}} = 77.94 \, \text{N}$

4. Calculate $\mu_k$: $\mu_k = \frac{F_{\text{friction}}}{N} = \frac{77.94}{200.25} = 0.39$

Final Answers:

1. Frictional force: $77.9 \, \text{N}$ (rounded)
2. Coefficient of kinetic friction: $0.39$

Would you like a more detailed explanation for any of these steps? Here are related questions to expand your understanding:

1. How does the angle of pulling affect the normal force?
2. Why does constant velocity imply that the net force is zero?
3. Can this method be applied if the surface is inclined? Why or why not?
4. What happens to the coefficient of friction if the angle is increased?
5. How does increasing the pulling force affect the object's motion?

Tip: Always resolve forces into components when dealing with inclined or angled forces for easier calculations.