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To find the horizontal acceleration of the block, we can apply Newton's second law \( F = ma \). We'll need to:

1. Break down the forces into their horizontal components.
2. Sum up all horizontal forces.
3. Use the total horizontal force to find the acceleration.

### Given:
- \( F_1 = 59.2 \, \text{N} \)
- \( F_2 = 20.5 \, \text{N} \)
- Mass of the block \( m = 7.00 \, \text{kg} \)
- Angle \( \theta = 70^\circ \) for \( F_1 \)
- Positive direction is to the right.

### Step 1: Resolve \( F_1 \) into components
Since \( F_1 \) is applied at an angle of 70° from the horizontal, we can resolve it into horizontal and vertical components:
- Horizontal component of \( F_1 \): 
  \[
  F_{1x} = F_1 \cos(70^\circ)
  \]
- Vertical component of \( F_1 \) (not relevant for horizontal acceleration): 
  \[
  F_{1y} = F_1 \sin(70^\circ)
  \]

### Step 2: Horizontal forces
- \( F_1 \) has a horizontal component \( F_{1x} \) to the right.
- \( F_2 \) is horizontal and points to the left (so it's negative when summing forces).

Thus, the total horizontal force is:
\[
F_{\text{total}} = F_{1x} - F_2
\]
Substituting the expressions:
\[
F_{\text{total}} = F_1 \cos(70^\circ) - F_2
\]

### Step 3: Apply Newton's second law to find acceleration
\[
F_{\text{total}} = ma
\]
\[
a = \frac{F_{\text{total}}}{m}
\]

Now, let's calculate the values.The horizontal acceleration of the block is approximately \( -0.036 \, \text{m/s}^2 \). The negative sign indicates that the acceleration is to the left, opposite to the positive direction defined as to the right.

Would you like more details or have any other questions?

Here are some related questions:
1. How does the vertical component of \( F_1 \) affect the system?
2. What would happen if the angle of \( F_1 \) were smaller?
3. How would the acceleration change if the block had more mass?
4. What are the forces acting on the block in the vertical direction?
5. How would the problem change if there were friction?

**Tip**: In problems with forces acting at angles, always resolve the forces into components before applying Newton's second law.

Two forces, F1 and F2, act on the 7.00-kg block shown in the drawing. The magnitudes of the forces are F1 = 59.2 N and F2 = 20.5 N. Take the positive direction to be to the right. Find the horizontal acceleration of the block, including sign.

Learn how to calculate the horizontal acceleration of a 7.00-kg block subjected to two forces, including one at an angle. Using Newton's Second Law and vector resolution, find the acceleration with given forces of F1 = 59.2 N and F2 = 20.5 N.

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