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What is the minimum mass mm that will stick and not slip?If this minimum mass is nudged ever so slightly, it will start being pulled up the incline. What acceleration will it have?

To analyze this problem, we need to break it down into two parts: determining the minimum mass \( m \) that will remain at rest (stick and not slip) and finding the acceleration if this mass starts moving. Let's go step by step.

### Part 1: Determining the Minimum Mass \( m \) that Will Stick and Not Slip
We have a mass \( m \) on an inclined plane at a 20° angle and another hanging mass of 2 kg. The system is connected by a rope passing over a pulley.

1. **For the hanging mass (2 kg):**
   - The force pulling the mass downward is its weight, \( F_{\text{gravity}} = m_{\text{hanging}} \cdot g = 2 \cdot 9.8 = 19.6 \, \text{N} \).
   
2. **For the block on the incline:**
   - The gravitational force acting on the block is \( F_{\text{gravity}} = m \cdot g \).
   - The force pulling the block down the incline is the component of gravity along the incline, \( F_{\text{down incline}} = m \cdot g \cdot \sin(20^\circ) \).
   - The normal force (perpendicular to the incline) is \( F_{\text{normal}} = m \cdot g \cdot \cos(20^\circ) \).
   
3. **Force balance for no slipping:**
   For the system to remain stationary (no slipping), the tension in the rope must balance the forces.
   
   - Tension \( T \) in the rope due to the hanging mass is equal to the weight of the hanging block: \( T = 19.6 \, \text{N} \).
   - For the block on the incline, the force down the incline due to gravity must be equal to this tension:
   \[
   m \cdot g \cdot \sin(20^\circ) = 19.6
   \]
   Substituting \( g = 9.8 \, \text{m/s}^2 \), we get:
   \[
   m \cdot 9.8 \cdot \sin(20^\circ) = 19.6
   \]
   \[
   m = \frac{19.6}{9.8 \cdot \sin(20^\circ)}
   \]
   Solving this gives:
   \[
   m = \frac{19.6}{9.8 \cdot 0.342} \approx 5.86 \, \text{kg}
   \]
   
So, the minimum mass \( m \) that will stick and not slip is approximately **5.86 kg**.

### Part 2: Acceleration When Nudged
If the block with mass \( m = 5.86 \, \text{kg} \) is nudged, it will begin to move. The net force acting on the system will now cause acceleration.

- The total mass of the system is the mass on the incline plus the hanging mass: \( m_{\text{total}} = 5.86 + 2 = 7.86 \, \text{kg} \).
- The net force is the difference between the gravitational force pulling the hanging mass down and the component of gravitational force pulling the block up the incline:
  \[
  F_{\text{net}} = 19.6 - (5.86 \cdot 9.8 \cdot \sin(20^\circ)) = 19.6 - 19.6 = 0 \, \text{N}
  \]
  
Since the net force is approximately zero, the system will not accelerate. Therefore, any slight movement due to nudging will not cause further acceleration, implying the system is in a state of equilibrium or has very negligible acceleration. This is a delicate case where even slight friction could play a role, but without friction specified, we assume the system doesn't accelerate.

Would you like further details on any step or have any questions? Here are some related questions for you to consider:

1. How does friction affect the minimum mass required for the system to stay at rest?
2. What happens if the angle of the incline is increased or decreased?
3. How would the result change if the mass of the hanging block is increased?
4. Can we generalize this result to different incline angles and hanging masses?
5. What would happen if the pulley had friction or mass?

**Tip:** Always account for the components of gravitational forces along and perpendicular to the incline in pulley problems involving inclined planes.

What is the minimum mass m that will stick and not slip? If this minimum mass is nudged ever so slightly, it will start being pulled up the incline. What acceleration will it have?

Determine the minimum mass required for a block on a 20-degree incline to stick without slipping and calculate the acceleration if nudged. Learn key concepts of pulley systems, inclined planes, and Newton's laws, with detailed calculations. Ideal for Grades 11-12 physics students.

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