College

Dear beloved readers, welcome to our website! We hope your visit here brings you valuable insights and meaningful inspiration. Thank you for taking the time to stop by and explore the content we've prepared for you.
------------------------------------------------ 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].