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the m of the inclined block is 1.9kg. If this minimum mass is nudged ever so slightly, it will start being pulled up the incline. What acceleration will it have?

To solve for the **acceleration** of the system shown in the image, we can model the forces acting on both the inclined block (mass \(m = 1.9 \, \text{kg}\)) and the hanging block (\(2 \, \text{kg}\)). Since the question states that nudging the system will pull the \(1.9 \, \text{kg}\) block **up the incline**, the net force will cause both masses to accelerate.

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### Step-by-Step Solution:

1. **Set up variables and assumptions:**
   - Mass of block on incline: \(m_1 = 1.9 \, \text{kg}\)
   - Mass of hanging block: \(m_2 = 2 \, \text{kg}\)
   - Angle of incline: \(\theta = 20^\circ\)
   - Gravitational acceleration: \(g = 9.8 \, \text{m/s}^2\)
   - Acceleration of the system: \(a\) (what we are solving for)
   - Assume a **frictionless** incline.

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### 2. Forces on each block:
- **Hanging mass** (\(m_2\)):  
  The tension \(T\) in the rope pulls it **upward**, while gravity pulls it **downward**:
  \[ m_2 g - T = m_2 a \tag{1} \]

- **Inclined block** (\(m_1\)):  
  The forces acting on this block are:
  - Component of gravitational force pulling it **down the incline**:  
    \(m_1 g \sin \theta\)
  - Tension \(T\) in the rope pulling it **up the incline**.

  So the equation of motion for this block is:
  \[ T - m_1 g \sin \theta = m_1 a \tag{2} \]

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### 3. Solve the system of equations:
From Equation (1):
\[
T = m_2 g - m_2 a
\]
Now, substitute this into Equation (2):
\[
m_2 g - m_2 a - m_1 g \sin \theta = m_1 a
\]
Rearrange terms:
\[
m_2 g - m_1 g \sin \theta = m_1 a + m_2 a
\]
Factor out \(a\) on the right:
\[
m_2 g - m_1 g \sin \theta = (m_1 + m_2) a
\]
Solve for \(a\):
\[
a = \frac{m_2 g - m_1 g \sin \theta}{m_1 + m_2}
\]

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### 4. Plug in the values:
- \(m_1 = 1.9 \, \text{kg}\)
- \(m_2 = 2 \, \text{kg}\)
- \(g = 9.8 \, \text{m/s}^2\)
- \(\sin 20^\circ \approx 0.342\)

\[
a = \frac{(2)(9.8) - (1.9)(9.8)(0.342)}{1.9 + 2}
\]

Evaluate step-by-step:
- \(2 \times 9.8 = 19.6\)
- \(1.9 \times 9.8 \times 0.342 \approx 6.37\)
- \(19.6 - 6.37 \approx 13.23\)
- \(1.9 + 2 = 3.9\)

\[
a = \frac{13.23}{3.9} \approx 3.39 \, \text{m/s}^2
\]

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### 5. Final Answer:
The **acceleration** of the system will be approximately:
\[
a \approx 3.39 \, \text{m/s}^2
\]

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Do you need any further details or clarifications? 

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Here are **5 related questions** to deepen your understanding:
1. How would the acceleration change if the mass on the incline was increased?
2. What happens if friction is introduced on the incline?
3. How does the tension in the rope compare to the force of gravity on the hanging mass?
4. What role does the angle of the incline play in the acceleration?
5. How would the system behave if the two masses were equal?

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**Tip:** When dealing with inclined planes, always break forces into components parallel and perpendicular to the surface for easier calculations.

The m of the inclined block is 1.9kg. If this minimum mass is nudged ever so slightly, it will start being pulled up the incline. What acceleration will it have?

Calculate the acceleration of a 1.9 kg block on a 20-degree incline connected to a 2 kg hanging block. Using Newton's second law, we analyze the forces acting on the system. Learn how to solve for acceleration step-by-step.

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