6. A spring gun has a constant of K= 650N/m. It is compressed a distance x 0.04m and a ball of mass 0.20kg is placed in the barrel against the compressed spring. The ball is originally at rest. a) Compute the maximum speed with which the ball leaves the gun when released and the spring is no longer compressed.

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**Spring Gun Physics Problem**

**Problem Statement:**

A spring gun has a constant of \( K = 650 \, \text{N/m} \). It is compressed a distance \( x = 0.04 \, \text{m} \) and a ball of mass \( 0.20 \, \text{kg} \) is placed in the barrel against the compressed spring. The ball is originally at rest.

**Tasks:**

a) Compute the maximum speed with which the ball leaves the gun when released and the spring is no longer compressed.

To solve this problem, we need to consider the conservation of energy principles. Specifically, the potential energy stored in the compressed spring will convert into the kinetic energy of the ball as it leaves the gun.

**Solution:**

1. **Potential Energy in Spring**:
   The potential energy \( U \) stored in a compressed spring is given by:
   \[
   U = \frac{1}{2} K x^2
   \]
   Substituting the given values:
   \[
   U = \frac{1}{2} \times 650 \, \text{N/m} \times (0.04 \, \text{m})^2
   \]
   \[
   U = \frac{1}{2} \times 650 \times 0.0016
   \]
   \[
   U = 0.52 \, \text{J}
   \]

2. **Kinetic Energy**:
   The kinetic energy \( K.E \) of the ball when it leaves the gun is:
   \[
   K.E = \frac{1}{2} mv^2
   \]
   where \( m \) is the mass of the ball and \( v \) is its velocity. 

   Since all the potential energy stored in the spring is converted into kinetic energy:
   \[
   0.52 \, \text{J} = \frac{1}{2} \times 0.20 \, \text{kg} \times v^2
   \]

3. **Solving for Velocity**:
   \[
   0.52 = 0.1 v^2
   \]
   \[
   v^2 = \frac{0.52}{0.1}
   \]
   \[
   v^2 = 5
Transcribed Image Text:**Spring Gun Physics Problem** **Problem Statement:** A spring gun has a constant of \( K = 650 \, \text{N/m} \). It is compressed a distance \( x = 0.04 \, \text{m} \) and a ball of mass \( 0.20 \, \text{kg} \) is placed in the barrel against the compressed spring. The ball is originally at rest. **Tasks:** a) Compute the maximum speed with which the ball leaves the gun when released and the spring is no longer compressed. To solve this problem, we need to consider the conservation of energy principles. Specifically, the potential energy stored in the compressed spring will convert into the kinetic energy of the ball as it leaves the gun. **Solution:** 1. **Potential Energy in Spring**: The potential energy \( U \) stored in a compressed spring is given by: \[ U = \frac{1}{2} K x^2 \] Substituting the given values: \[ U = \frac{1}{2} \times 650 \, \text{N/m} \times (0.04 \, \text{m})^2 \] \[ U = \frac{1}{2} \times 650 \times 0.0016 \] \[ U = 0.52 \, \text{J} \] 2. **Kinetic Energy**: The kinetic energy \( K.E \) of the ball when it leaves the gun is: \[ K.E = \frac{1}{2} mv^2 \] where \( m \) is the mass of the ball and \( v \) is its velocity. Since all the potential energy stored in the spring is converted into kinetic energy: \[ 0.52 \, \text{J} = \frac{1}{2} \times 0.20 \, \text{kg} \times v^2 \] 3. **Solving for Velocity**: \[ 0.52 = 0.1 v^2 \] \[ v^2 = \frac{0.52}{0.1} \] \[ v^2 = 5
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