The equations below are the hori-
zontal x - and vertical y - component forms of Newton’s

second law applied to a physical process. Solve for the un-
knowns. Then work backward and construct a force diagram

for the object of interest and invent a problem for which the
equations might be an answer (there are many possibilities).
15.0 kg2ax = 150 N2cos 30 + N cos 90
+ 15.0 kg219.8 N>kg2cos 90
15.0 kg20 = 1-50 N2 sin 30 + N sin 90
- 15.0 kg219.8 N>kg2sin 90

Let's solve the given equations step by step.

The equations provided are:

1. \( 15.0 \, \text{kg} \cdot a_x = 150 \, \text{N} \cdot \cos(30^\circ) + N \cdot \cos(90^\circ) + 15.0 \, \text{kg} \cdot 19.8 \, \text{N/kg} \cdot \cos(90^\circ) \)
2. \( 15.0 \, \text{kg} \cdot 0 = 150 \, \text{N} \cdot \sin(30^\circ) + N \cdot \sin(90^\circ) - 15.0 \, \text{kg} \cdot 19.8 \, \text{N/kg} \cdot \sin(90^\circ) \)

### Step 1: Simplify the equations

**Equation 1: Horizontal Component**

First, note that:
- \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \)
- \( \cos(90^\circ) = 0 \)
- \( \cos(90^\circ) = 0 \)

Thus, equation 1 simplifies to:
\[
15.0 \, a_x = 150 \cdot \frac{\sqrt{3}}{2}
\]

Calculating \( 150 \cdot \frac{\sqrt{3}}{2} \):
\[
150 \cdot \frac{\sqrt{3}}{2} \approx 150 \cdot 0.866 \approx 129.9 \, \text{N}
\]

Therefore, we have:
\[
15.0 \, a_x = 129.9 \implies a_x = \frac{129.9}{15.0} \approx 8.66 \, \text{m/s}^2
\]

**Equation 2: Vertical Component**

Now simplify equation 2:
- \( \sin(30^\circ) = \frac{1}{2} \)
- \( \sin(90^\circ) = 1 \)

Thus, equation 2 simplifies to:
\[
0 = 150 \cdot \frac{1}{2} + N - 15.0 \cdot 19.8
\]
\[
0 = 75 + N - 297
\]
\[
N = 297 - 75 = 222 \, \text{N}
\]

### Summary of Results

- **Acceleration in the x-direction**: \( a_x \approx 8.66 \, \text{m/s}^2 \)
- **Normal Force \( N \)**: \( N = 222 \, \text{N} \)

### Step 2: Force Diagram

1. **Weight Force (\( W \))**: Acts downwards, \( W = 15.0 \, \text{kg} \cdot 19.8 \, \text{N/kg} = 297 \, \text{N} \).
2. **Normal Force (\( N \))**: Acts upwards, \( N = 222 \, \text{N} \).
3. **Applied Force (\( F_a \))**: \( 150 \, \text{N} \) at \( 30^\circ \) to the horizontal.
4. **Net Force**: Resultant of horizontal forces providing \( a_x \).

### Invented Problem

**Problem Scenario**: A box weighing 15.0 kg is on a flat surface. An external force of 150 N is applied at an angle of 30 degrees from the horizontal. Determine the box's acceleration and the normal force acting on it, considering the gravitational force. 

This scenario uses the same physics principles represented in the equations.

### Questions for Further Exploration

1. How does the angle of the applied force affect the acceleration of the box?
2. What would happen to the normal force if the box were on an incline?
3. How would the situation change if the mass of the box were increased?
4. Can you calculate the frictional force acting on the box if the coefficient of friction is known?
5. What are the implications of the net force being zero in terms of the box's motion?

### Tip
When working with forces, always draw a free-body diagram to visualize all forces acting on the object; it aids in understanding the problem better. Would you like more details on any part of this solution?

The equations below are the horizontal x - and vertical y - component forms of Newton’s second law applied to a physical process. Solve for the unknowns. Then work backward and construct a force diagram for the object of interest and invent a problem for which the equations might be an answer.

Explore how to solve Newton's second law equations and construct force diagrams in this detailed guide. We cover acceleration calculations and normal force determination for a box under an applied force of 150 N at an angle of 30 degrees. Ideal for high school physics students.

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