To solve for the magnitude of acceleration $a$ of the block, we need to analyze the forces acting on it and apply Newton's Second Law.

Step 1: Analyze the Forces

• Force $F_1$: This is a horizontal force of 55 N.

• Force $F_2$: This force has a magnitude of 10.3 N and is applied at an angle $\theta = 30^\circ$ to the horizontal. We can break this force into its horizontal and vertical components:

  • Horizontal component of $F_2$: $F_{2x} = F_2 \cdot \cos(\theta)$
  • Vertical component of $F_2$: $F_{2y} = F_2 \cdot \sin(\theta)$

• Frictional Force: The frictional force $F_{\text{friction}}$ opposes the motion and can be calculated using the coefficient of kinetic friction $\mu_k = 0.2$ and the normal force $F_N$ (which was previously calculated as 41.45 N): $F_{\text{friction}} = \mu_k \cdot F_N$

• Normal Force $F_N$: The normal force was already determined as 41.45 N in part (a).

Step 2: Newton’s Second Law

To find the acceleration, we apply Newton’s Second Law in the horizontal direction. The forces in the horizontal direction are:

• $F_1$
• The horizontal component of $F_2$, $F_{2x}$
• The frictional force, $F_{\text{friction}}$

The net force in the horizontal direction is: $F_{\text{net}} = F_1 + F_{2x} - F_{\text{friction}}$

Using Newton’s Second Law: $F_{\text{net}} = m \cdot a$ Thus, the equation for acceleration is: $a = \frac{F_{\text{net}}}{m}$

Step 3: Calculation

Let’s calculate each of the components:

• Horizontal component of $F_2$: $F_{2x} = 10.3 \, \text{N} \cdot \cos(30^\circ) = 10.3 \, \text{N} \cdot 0.866 = 8.91 \, \text{N}$

• Frictional force: $F_{\text{friction}} = 0.2 \cdot 41.45 \, \text{N} = 8.29 \, \text{N}$

Now, we can find the net force: $F_{\text{net}} = 55 \, \text{N} + 8.91 \, \text{N} - 8.29 \, \text{N} = 55 \, \text{N} + 0.62 \, \text{N} = 55.62 \, \text{N}$

Finally, using Newton’s Second Law to find the acceleration: $a = \frac{55.62 \, \text{N}}{3.7 \, \text{kg}} = 15.03 \, \text{m/s}^2$

Final Answer:

The magnitude of the acceleration is approximately: $a = 15.03 \, \text{m/s}^2$

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Would you like further details or clarifications on this solution? Here are a few related questions you could explore:

1. How would the acceleration change if the angle $\theta$ were increased to 45°?
2. What would happen to the normal force if the coefficient of friction were increased to 0.3?
3. How can we calculate the work done by the frictional force on the block?
4. What would the acceleration be if the block had a different mass, say 5 kg?
5. How can the net force be affected if the direction of $F_2$ is reversed?

Tip: When calculating frictional forces, always ensure you're using the correct normal force, as it directly influences the frictional interaction between surfaces.