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This problem involves the motion of a spring-mass system. Let's break down the key information:

### Given:
1. **Force stretching the spring**: \( 960 \, \text{N} \)
2. **Spring stretches**: \( 4 \, \text{m} \)
3. **Mass of the object**: \( 60 \, \text{kg} \)
4. **Initial position**: \( x(0) = 0 \, \text{m} \) (the mass is released from equilibrium)
5. **Initial velocity**: \( x'(0) = -10 \, \text{m/s} \) (velocity is upward)

### Key Objective:
Find the **equation of motion** for the system, which is given as:
\[
x(t) = -5 \sin(2t) \, \text{m}
\]

### Step-by-Step Breakdown:
1. **Find the Spring Constant (k)**:
   Using Hooke's law \( F = kx \), where:
   \[
   F = 960 \, \text{N}, \quad x = 4 \, \text{m}
   \]
   Solving for \( k \):
   \[
   k = \frac{F}{x} = \frac{960}{4} = 240 \, \text{N/m}
   \]

2. **Determine the Angular Frequency (\( \omega \))**:
   The angular frequency \( \omega \) of a spring-mass system is given by:
   \[
   \omega = \sqrt{\frac{k}{m}}
   \]
   Substituting \( k = 240 \, \text{N/m} \) and \( m = 60 \, \text{kg} \):
   \[
   \omega = \sqrt{\frac{240}{60}} = \sqrt{4} = 2 \, \text{rad/s}
   \]

3. **Initial Conditions and General Solution**:
   The general solution for a spring-mass system is:
   \[
   x(t) = A \cos(\omega t) + B \sin(\omega t)
   \]
   From the initial conditions \( x(0) = 0 \), we get:
   \[
   0 = A \cos(0) + B \sin(0) \quad \Rightarrow \quad A = 0
   \]
   Now the equation simplifies to:
   \[
   x(t) = B \sin(2t)
   \]
   Using the initial velocity condition \( x'(0) = -10 \, \text{m/s} \):
   \[
   x'(t) = 2B \cos(2t)
   \]
   At \( t = 0 \):
   \[
   x'(0) = 2B \cos(0) = 2B
   \]
   Setting this equal to the initial velocity \( -10 \):
   \[
   2B = -10 \quad \Rightarrow \quad B = -5
   \]

4. **Final Equation of Motion**:
   Therefore, the equation of motion is:
   \[
   x(t) = -5 \sin(2t) \, \text{m}
   \]

This confirms the given solution.

Would you like further clarification on any part of this? Here are a few related questions to extend this concept:

1. How would the equation of motion change if the mass were doubled?
2. What effect does increasing the spring constant have on the frequency of the system?
3. How does damping affect the motion of the spring?
4. Can you calculate the total mechanical energy of the system?
5. How would you model this system with an external force acting on it?

**Tip**: The angular frequency \( \omega \) is directly influenced by the spring constant and the mass. A higher spring constant or a lower mass will increase the oscillation frequency.

A force of 960 newtons stretches a spring 4 meters. A mass of 60 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 10 m/s. Find the equation of motion.

This problem involves a spring-mass system where a force of 960 N stretches a spring by 4 meters. A mass of 60 kg is attached and released with an initial velocity of 10 m/s. The solution covers concepts such as Hooke's Law, angular frequency, and solving differential equations.

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