Note that when the system is at rest in the equilibrium position, the spring is already stretched. I don't want question 19 from the attached photo answered. Just this one: 20. Write an expression for how far the spring is stretched when the cart is at rest in the equilibriumposition. Name this distance y0.In the data run, the positions are measured from the rest (equilibrium) position of the cart. Name thisposition y1 . The distance that the spring is stretched is y=y0 - y1. The minus sign is because apositive position for the cart decreases the spring stretch. (See the attached image.) Substitute your expression from question 20 into y=y0-y1. **Part C – Newton’s Second Law and the Simple Harmonic Oscillator**In this part, you will explore how Newton’s Second Law relates to Simple Harmonic Motion and the ideal spring. The spring pulls to the left on the cart, and gravity pulls down on the hanging mass. As a result, the cart accelerates back and forth, and the mass accelerates up and down. The force sensor is zeroed at the cart’s equilibrium position, measuring how much more or less than the spring force at this position.### Diagram Explanation- The diagram shows a spring with force constant \( k \) attached to a cart with mass \( m_c \).- A force sensor is placed at the equilibrium position of the cart, reading zero when the system is at equilibrium.- The cart's position changes from \( y_0 \) (when the spring is unstretched) to \( y_1 \) (current position).- A hanging mass \( m_h \) is shown at the equilibrium position of the mass, with its position changing similarly from \( y_0 \) to \( y_1 \).### Analysis**19. Analyze this as a physics 4A-style force problem:**- Define the distance the spring is stretched as \( y \).- Specify the mass of the cart (\( m_c \)) and the mass of the hanger (\( m_h \)).- Solve for the acceleration, \( a \), as a function of spring stretch.- Note: The cart measures positive direction to the left, so positive acceleration is leftward.**Note:** The system is at rest at the equilibrium position; the spring is pre-stretched.

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Description: Note that when the system is at rest in the equilibrium position, the spring is already stretched. I don't want question 19 from the attached photo answered. Just this one: 20. Write an expression for how far the spring is stretched when the c

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Note that when the system is at rest in the equilibrium position, the spring is already stretched.

I don't want question 19 from the attached photo answered. Just this one:

20. Write an expression for how far the spring is stretched when the cart is at rest in the equilibrium
position. Name this distance y_0.

_{In the data run, the positions are measured from the rest (equilibrium) position of the cart. Name this
position} y₁ _{. The distance that the spring is stretched is y=y0 - y1. The minus sign is because a
positive position for the cart decreases the spring stretch. (See the attached image.) Substitute your expression from question 20 into y=y0-y1.}

**Part C – Newton’s Second Law and the Simple Harmonic Oscillator**

In this part, you will explore how Newton’s Second Law relates to Simple Harmonic Motion and the ideal spring. The spring pulls to the left on the cart, and gravity pulls down on the hanging mass. As a result, the cart accelerates back and forth, and the mass accelerates up and down. The force sensor is zeroed at the cart’s equilibrium position, measuring how much more or less than the spring force at this position.

### Diagram Explanation

- The diagram shows a spring with force constant \( k \) attached to a cart with mass \( m_c \).
- A force sensor is placed at the equilibrium position of the cart, reading zero when the system is at equilibrium.
- The cart's position changes from \( y_0 \) (when the spring is unstretched) to \( y_1 \) (current position).
- A hanging mass \( m_h \) is shown at the equilibrium position of the mass, with its position changing similarly from \( y_0 \) to \( y_1 \).

### Analysis

**19. Analyze this as a physics 4A-style force problem:**
- Define the distance the spring is stretched as \( y \).
- Specify the mass of the cart (\( m_c \)) and the mass of the hanger (\( m_h \)).
- Solve for the acceleration, \( a \), as a function of spring stretch.
- Note: The cart measures positive direction to the left, so positive acceleration is leftward.

**Note:** The system is at rest at the equilibrium position; the spring is pre-stretched.

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