To determine the horizontal acceleration of the block, we need to analyze the forces involved and use Newton's second law. Here’s a step-by-step explanation:
1. Identification of the Given Data:
- Mass of the block, [tex]\( m = 5 \, \text{kg} \)[/tex]
- Pulling force, [tex]\( F = 26.43 \, \text{N} \)[/tex]
- Angle of the pulling force, [tex]\( \theta = 30^\circ \)[/tex]
- Coefficient of kinetic friction, [tex]\( \mu_k = 0.11 \)[/tex]
- Acceleration due to gravity, [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex]
2. Convert the Angle to Radians:
- [tex]\(\theta \)[/tex] in radians is [tex]\( \theta = \frac{\pi}{6} \)[/tex] or [tex]\( \theta \approx 0.5236 \)[/tex] radians.
3. Resolve the Pulling Force into Horizontal and Vertical Components:
- Horizontal component of the force: [tex]\( F_{\text{horizontal}} = F \cos(\theta) \)[/tex]
[tex]\[
F_{\text{horizontal}} = 26.43 \times \cos(30^\circ) \approx 22.89 \, \text{N}
\][/tex]
- Vertical component of the force: [tex]\( F_{\text{vertical}} = F \sin(\theta) \)[/tex]
[tex]\[
F_{\text{vertical}} = 26.43 \times \sin(30^\circ) \approx 13.21 \, \text{N}
\][/tex]
4. Calculate the Gravitational Force (Weight) and Normal Force:
- Gravitational force: [tex]\( F_{\text{gravity}} = m \times g \)[/tex]
[tex]\[
F_{\text{gravity}} = 5 \times 9.81 \approx 49.05 \, \text{N}
\][/tex]
- Normal force: [tex]\( F_{\text{normal}} = F_{\text{gravity}} - F_{\text{vertical}} \)[/tex]
[tex]\[
F_{\text{normal}} = 49.05 - 13.21 \approx 35.84 \, \text{N}
\][/tex]
5. Calculate the Frictional Force:
- Frictional force: [tex]\( F_{\text{friction}} = \mu_k \times F_{\text{normal}} \)[/tex]
[tex]\[
F_{\text{friction}} = 0.11 \times 35.84 \approx 3.94 \, \text{N}
\][/tex]
6. Determine the Net Horizontal Force Acting on the Block:
- Net horizontal force: [tex]\( F_{\text{net horizontal}} = F_{\text{horizontal}} - F_{\text{friction}} \)[/tex]
[tex]\[
F_{\text{net horizontal}} = 22.89 - 3.94 \approx 18.95 \, \text{N}
\][/tex]
7. Calculate the Horizontal Acceleration of the Block:
- Using Newton's second law: [tex]\( F = m \times a \)[/tex]
[tex]\[
a = \frac{F_{\text{net horizontal}}}{m}
\][/tex]
[tex]\[
a = \frac{18.95}{5} \approx 3.79 \, \text{m/s}^2
\][/tex]
Therefore, the horizontal acceleration of the block is approximately [tex]\( 3.79 \, \text{m/s}^2 \)[/tex].
