Question
A 45-kg block is being pulled up a frictionless inclined plane at an angle of 30 degrees, with a force of 120 N applied parallel to the surface. What is the resulting acceleration of the block?
Answers
GPT-4o mini
To determine the acceleration of the block being pulled up the frictionless inclined plane, we can follow these steps:
1. **Determine the gravitational force acting on the block:**
The weight of the block \( W \) can be calculated using the formula:
\[
W = m \cdot g
\]
where \( m = 45 \, \text{kg} \) (mass of the block) and \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity).
\[
W = 45 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 441.45 \, \text{N}
\]
2. **Calculate the component of the gravitational force acting down the incline:**
The component of the weight acting down the incline can be calculated as:
\[
W_{\text{down}} = W \cdot \sin(\theta)
\]
where \( \theta = 30^\circ \). Thus,
\[
W_{\text{down}} = 441.45 \, \text{N} \cdot \sin(30^\circ) = 441.45 \, \text{N} \cdot 0.5 = 220.725 \, \text{N}
\]
3. **Set up the net force equation:**
The net force \( F_{\text{net}} \) acting on the block can be found by taking the applied force and subtracting the gravitational force component acting down the incline:
\[
F_{\text{net}} = F_{\text{applied}} - W_{\text{down}}
\]
where \( F_{\text{applied}} = 120 \, \text{N} \).
Thus,
\[
F_{\text{net}} = 120 \, \text{N} - 220.725 \, \text{N} = -100.725 \, \text{N}
\]
4. **Calculate the acceleration:**
The net force can then be used to find the acceleration \( a \) using Newton’s second law:
\[
F_{\text{net}} = m \cdot a
\]
Rearranging for \( a \):
\[
a = \frac{F_{\text{net}}}{m} = \frac{-100.725 \, \text{N}}{45 \, \text{kg}} \approx -2.24 \, \text{m/s}^2
\]
The negative sign indicates that the acceleration is directed down the incline, suggesting that the applied force is not sufficient to overcome the component of gravity pulling the block down.
**Final Result:**
The resulting acceleration of the block is approximately **-2.24 m/s²** (indicating the block accelerates down the incline).
1. **Determine the gravitational force acting on the block:**
The weight of the block \( W \) can be calculated using the formula:
\[
W = m \cdot g
\]
where \( m = 45 \, \text{kg} \) (mass of the block) and \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity).
\[
W = 45 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 441.45 \, \text{N}
\]
2. **Calculate the component of the gravitational force acting down the incline:**
The component of the weight acting down the incline can be calculated as:
\[
W_{\text{down}} = W \cdot \sin(\theta)
\]
where \( \theta = 30^\circ \). Thus,
\[
W_{\text{down}} = 441.45 \, \text{N} \cdot \sin(30^\circ) = 441.45 \, \text{N} \cdot 0.5 = 220.725 \, \text{N}
\]
3. **Set up the net force equation:**
The net force \( F_{\text{net}} \) acting on the block can be found by taking the applied force and subtracting the gravitational force component acting down the incline:
\[
F_{\text{net}} = F_{\text{applied}} - W_{\text{down}}
\]
where \( F_{\text{applied}} = 120 \, \text{N} \).
Thus,
\[
F_{\text{net}} = 120 \, \text{N} - 220.725 \, \text{N} = -100.725 \, \text{N}
\]
4. **Calculate the acceleration:**
The net force can then be used to find the acceleration \( a \) using Newton’s second law:
\[
F_{\text{net}} = m \cdot a
\]
Rearranging for \( a \):
\[
a = \frac{F_{\text{net}}}{m} = \frac{-100.725 \, \text{N}}{45 \, \text{kg}} \approx -2.24 \, \text{m/s}^2
\]
The negative sign indicates that the acceleration is directed down the incline, suggesting that the applied force is not sufficient to overcome the component of gravity pulling the block down.
**Final Result:**
The resulting acceleration of the block is approximately **-2.24 m/s²** (indicating the block accelerates down the incline).