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We are tasked with determining the magnitude and direction of the puck's acceleration. To do this, we'll analyze the forces acting on the puck and use Newton's second law of motion: \( \vec{F}_{\text{net}} = m \cdot \vec{a} \), where:

- \( m = 0.30 \, \text{kg} \) is the mass of the puck,
- \( \vec{F}_{\text{net}} \) is the vector sum of the two applied forces.

The forces provided are:
- \( F_1 = 5.0 \, \text{N} \) at an angle of \( 20^\circ \) below the x-axis,
- \( F_2 = 8.0 \, \text{N} \) at an angle of \( 60^\circ \) above the x-axis.

### Step 1: Break Forces into Components

We'll resolve each force into its horizontal (x) and vertical (y) components.

- For \( F_1 \) (with angle \( \theta_1 = 20^\circ \)):
  \[
  F_{1x} = F_1 \cos(20^\circ) = 5.0 \cos(20^\circ)
  \]
  \[
  F_{1y} = F_1 \sin(20^\circ) = -5.0 \sin(20^\circ) \quad \text{(negative since it is below the x-axis)}
  \]

- For \( F_2 \) (with angle \( \theta_2 = 60^\circ \)):
  \[
  F_{2x} = F_2 \cos(60^\circ) = 8.0 \cos(60^\circ)
  \]
  \[
  F_{2y} = F_2 \sin(60^\circ) = 8.0 \sin(60^\circ)
  \]

### Step 2: Sum the Components

Now, sum the x and y components of the forces to find the net force.

- Net force in the x-direction:
  \[
  F_{\text{net},x} = F_{1x} + F_{2x}
  \]

- Net force in the y-direction:
  \[
  F_{\text{net},y} = F_{1y} + F_{2y}
  \]

### Step 3: Calculate the Magnitude of the Net Force

The magnitude of the net force is given by:
\[
F_{\text{net}} = \sqrt{(F_{\text{net},x})^2 + (F_{\text{net},y})^2}
\]

### Step 4: Calculate the Direction of the Net Force

The direction \( \theta_{\text{net}} \) of the net force relative to the positive x-axis is:
\[
\theta_{\text{net}} = \tan^{-1} \left( \frac{F_{\text{net},y}}{F_{\text{net},x}} \right)
\]

### Step 5: Use Newton's Second Law to Find Acceleration

Finally, use Newton's second law to find the magnitude of the acceleration:
\[
a = \frac{F_{\text{net}}}{m}
\]

---

Let me perform the calculations next.Here are the results of the calculations:

- The net force components are:
  - \( F_{\text{net},x} = 8.70 \, \text{N} \)
  - \( F_{\text{net},y} = 5.22 \, \text{N} \)

- The magnitude of the net force is \( F_{\text{net}} = 10.14 \, \text{N} \).

- The direction of the net force is \( \theta_{\text{net}} = 30.96^\circ \) above the positive x-axis.

- The magnitude of the puck's acceleration is \( a = 33.81 \, \text{m/s}^2 \).

Would you like more details on any of these steps or further clarification?

Here are five relative questions for further practice:
1. How would the acceleration change if the puck's mass was doubled?
2. What would happen to the direction of the net force if \( F_2 \) was increased to 10 N?
3. How would the problem change if friction was introduced to the ice?
4. How does Newton's second law relate to the calculation of acceleration in this scenario?
5. What would be the new net force if the angle of \( F_1 \) was \( 30^\circ \) instead of \( 20^\circ \)?

**Tip**: When calculating forces, always resolve them into components to simplify the vector addition process.

A hockey puck having a mass of 0.30 kg slides on a frictionless ice rink. Two hockey sticks strike the puck simultaneously, exerting forces F1 = 5.0 N at an angle of 20° below the x-axis, and F2 = 8.0 N at an angle of 60° above the x-axis. Determine both the magnitude and direction of the puck's acceleration.

This physics problem involves calculating the acceleration of a hockey puck on a frictionless surface. The problem explores vector addition, force components, and Newton's second law. Learn how to break down forces and find the resulting acceleration. Suitable for high school students in Grades 9-12.

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