uppose you fire a mass m straight vertically upwards from a spring that is knc compressed a distance x from its equilibrium position, and the mass reac eight of h above its launch point. Using the conservation of energy, derive

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### Problem Statement

Suppose you fire a mass \( m \) straight vertically upwards from a spring that is known to have been compressed a distance \( x \) from its equilibrium position, and the mass reaches a maximum height of \( h \) above its launch point. Using the conservation of energy, derive an expression for the spring constant \( k \) of the spring.

### Educational Explanation

To solve this problem, we apply the principle of conservation of energy, which states that the total mechanical energy in the system remains constant if only conservative forces act on it.

#### 1. Initial Energy
Initially, the spring is compressed by a distance \( x \). The potential energy stored in the spring, known as elastic potential energy, can be expressed as:
\[ 
E_{\text{spring}} = \frac{1}{2}kx^2 
\]
where \( k \) is the spring constant.

#### 2. Energy at Maximum Height
At the maximum height \( h \), all the initial spring potential energy is converted into gravitational potential energy. The gravitational potential energy is given by:
\[ 
E_{\text{gravity}} = mgh 
\]
where \( g \) is the acceleration due to gravity.

#### 3. Conservation of Energy
According to the conservation of energy:
\[ 
\frac{1}{2}kx^2 = mgh 
\]

#### 4. Deriving the Spring Constant
We need to solve for \( k \). Rearranging the equation:
\[ 
k = \frac{2mgh}{x^2} 
\]

### Conclusion
Thus, by using the conservation of energy, we derived the expression for the spring constant:
\[ 
k = \frac{2mgh}{x^2} 
\]
This expression shows how the spring constant is related to the mass, gravitational acceleration, maximum height, and compression distance.
Transcribed Image Text:### Problem Statement Suppose you fire a mass \( m \) straight vertically upwards from a spring that is known to have been compressed a distance \( x \) from its equilibrium position, and the mass reaches a maximum height of \( h \) above its launch point. Using the conservation of energy, derive an expression for the spring constant \( k \) of the spring. ### Educational Explanation To solve this problem, we apply the principle of conservation of energy, which states that the total mechanical energy in the system remains constant if only conservative forces act on it. #### 1. Initial Energy Initially, the spring is compressed by a distance \( x \). The potential energy stored in the spring, known as elastic potential energy, can be expressed as: \[ E_{\text{spring}} = \frac{1}{2}kx^2 \] where \( k \) is the spring constant. #### 2. Energy at Maximum Height At the maximum height \( h \), all the initial spring potential energy is converted into gravitational potential energy. The gravitational potential energy is given by: \[ E_{\text{gravity}} = mgh \] where \( g \) is the acceleration due to gravity. #### 3. Conservation of Energy According to the conservation of energy: \[ \frac{1}{2}kx^2 = mgh \] #### 4. Deriving the Spring Constant We need to solve for \( k \). Rearranging the equation: \[ k = \frac{2mgh}{x^2} \] ### Conclusion Thus, by using the conservation of energy, we derived the expression for the spring constant: \[ k = \frac{2mgh}{x^2} \] This expression shows how the spring constant is related to the mass, gravitational acceleration, maximum height, and compression distance.
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