College

A 305 kg block is pulled by two horizontal forces. The first force is 124 N at a [tex]$29.6^{\circ}$[/tex] angle, and the second is 187 N at a [tex]$194^{\circ}$[/tex] angle.

What is the acceleration of the block?

[tex]a = [?] \, \text{m/s}^2[/tex]

Answer :

To find the acceleration of the block, you need to consider the forces acting on it and use Newton's second law, which states that [tex]\( F_{\text{net}} = m \cdot a \)[/tex], where [tex]\( F_{\text{net}} \)[/tex] is the net force, [tex]\( m \)[/tex] is the mass, and [tex]\( a \)[/tex] is the acceleration.

Here's a step-by-step breakdown:

1. Identify Given Values:
- Mass of the block, [tex]\( m = 305 \, \text{kg} \)[/tex].
- First force, [tex]\( F_1 = 124 \, \text{N} \)[/tex] at an angle of [tex]\( 29.6^\circ \)[/tex].
- Second force, [tex]\( F_2 = 187 \, \text{N} \)[/tex] at an angle of [tex]\( 194^\circ \)[/tex].

2. Resolve Forces into Components:
- Convert angles from degrees to radians if needed (for calculations involving trigonometric functions).
- For [tex]\( F_1 \)[/tex]:
- [tex]\( F_{1x} = F_1 \cdot \cos(29.6^\circ) \)[/tex]
- [tex]\( F_{1y} = F_1 \cdot \sin(29.6^\circ) \)[/tex]

- For [tex]\( F_2 \)[/tex]:
- [tex]\( F_{2x} = F_2 \cdot \cos(194^\circ) \)[/tex]
- [tex]\( F_{2y} = F_2 \cdot \sin(194^\circ) \)[/tex]

3. Calculate Net Force Components:
- The x-component of the net force: [tex]\( F_{\text{net}_x} = F_{1x} + F_{2x} \)[/tex]
- The y-component of the net force: [tex]\( F_{\text{net}_y} = F_{1y} + F_{2y} \)[/tex]

4. Calculate the Magnitude of the Net Force:
- Use the Pythagorean theorem:
[tex]\[ F_{\text{net}} = \sqrt{F_{\text{net}_x}^2 + F_{\text{net}_y}^2} \][/tex]

5. Calculate Acceleration:
- Use Newton's second law to find acceleration:
[tex]\[ a = \frac{F_{\text{net}}}{m} \][/tex]

By following these steps based on the provided data, the magnitude of the net force is approximately [tex]\( 75.35 \, \text{N} \)[/tex] and the acceleration of the block is approximately [tex]\( 0.247 \, \text{m/s}^2 \)[/tex].