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Figuur 6 toont een mof glijdend over een gladde buis. De 3 kg zware mof is bevestigd aan een veer met een rustlengte van 2 m. Als de mof naar punt B wordt getrokken en vervolgens losgelaten vanuit rust, bepaal dan de grootte van zijn snelheid bij aankomst in punt A.

We are dealing with a system where a 3 kg mass (the "mof") is attached to a spring with a spring constant \( k = 3 \, \text{N/m} \). The spring has an unstretched length of 2 m. Initially, the mass is pulled to point B, which is 4 m horizontally from point A, and released from rest. We are asked to determine the velocity of the mass when it reaches point A (located 4 m horizontally and 3 m vertically from the initial position).

### Steps to Solve:
1. **Spring Energy (Elastic Potential Energy) at point B**:
   The spring is stretched initially from its natural length of 2 m. The total stretch at point B needs to be computed based on the distance between points A and B.

   - The distance between A and B, using the Pythagorean theorem:
     \[
     d_{AB} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{m}
     \]
   - The extension of the spring is the difference between this distance and the natural length of the spring:
     \[
     \Delta L = 5 - 2 = 3 \, \text{m}
     \]
   - The potential energy stored in the spring at point B is given by:
     \[
     PE_{spring} = \frac{1}{2} k (\Delta L)^2 = \frac{1}{2} \times 3 \times (3)^2 = \frac{1}{2} \times 3 \times 9 = 13.5 \, \text{J}
     \]

2. **Kinetic Energy at point A**:
   At point A, all the potential energy stored in the spring will have converted to kinetic energy (assuming no other losses like friction):
   \[
   KE = \frac{1}{2} m v^2
   \]
   Setting the kinetic energy equal to the potential energy from the spring at point B:
   \[
   \frac{1}{2} m v^2 = 13.5
   \]
   Plugging in the mass \( m = 3 \, \text{kg} \):
   \[
   \frac{1}{2} \times 3 \times v^2 = 13.5
   \]
   Solving for \( v^2 \):
   \[
   v^2 = \frac{13.5 \times 2}{3} = 9 \quad \Rightarrow \quad v = \sqrt{9} = 3 \, \text{m/s}
   \]

Thus, the speed of the mass at point A is **3 m/s**.

Would you like further details on the spring potential energy or kinetic energy concepts?

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### 5 Related Questions:
1. What happens if the spring constant \( k \) is doubled? How will the velocity at point A change?
2. How would the speed change if the mass of the "mof" were 6 kg instead of 3 kg?
3. What role does the gravitational potential energy play in this problem, if any?
4. How does the energy conservation principle apply when friction is present on the surface?
5. What if the spring was compressed instead of stretched initially—how would that affect the system?

### Tip:
Always use the principle of energy conservation in mechanical systems without friction to simplify complex problems involving springs and motion.

Solve a physics problem involving a 3 kg mass attached to a spring with a spring constant of 3 N/m and a natural length of 2 meters. The mass is pulled to a point 4 meters horizontally and 3 meters vertically from the rest position and released. Learn how to use energy conservation principles to determine the velocity at point A.

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