Question

A 5 kg block is pulled up a frictionless inclined plane with a force of 30 N. If the angle of the plane is 25 degrees, what is the acceleration of the block?

Answer 1

To find the acceleration of the block being pulled up the inclined plane, we can start by analyzing the forces acting on the block.

1. **Forces acting on the block**:
- The gravitational force acting downwards: \( F_g = m \cdot g \) where \( m = 5 \) kg and \( g \approx 9.81 \, \text{m/s}^2 \).
- The pulling force \( F_{\text{applied}} = 30 \) N acting up the incline.
- The component of the gravitational force acting down the incline: \( F_{\text{gravity, parallel}} = F_g \cdot \sin(\theta) \).

2. **Calculating the gravitational force**:
\[
F_g = m \cdot g = 5 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 49.05 \, \text{N}
\]

3. **Calculating the component of the gravitational force along the incline**:
\[
F_{\text{gravity, parallel}} = F_g \cdot \sin(25^\circ)
\]
\[
F_{\text{gravity, parallel}} = 49.05 \, \text{N} \cdot \sin(25^\circ)
\]
Using \( \sin(25^\circ) \approx 0.4226 \):
\[
F_{\text{gravity, parallel}} = 49.05 \, \text{N} \cdot 0.4226 \approx 20.74 \, \text{N}
\]

4. **Calculating the net force \( F_{\text{net}} \)** acting on the block:
The net force is the pulling force minus the gravitational force component acting down the incline:
\[
F_{\text{net}} = F_{\text{applied}} - F_{\text{gravity, parallel}}
\]
\[
F_{\text{net}} = 30 \, \text{N} - 20.74 \, \text{N} \approx 9.26 \, \text{N}
\]

5. **Calculating the acceleration \( a \)** using Newton's second law \( F = m \cdot a \):
\[
a = \frac{F_{\text{net}}}{m} = \frac{9.26 \, \text{N}}{5 \, \text{kg}} \approx 1.852 \, \text{m/s}^2
\]

Thus, the acceleration of the block is approximately \( \mathbf{1.85 \, \text{m/s}^2} \).