Answer :
Let's compare the potential energies of both the object and the spring after doubling the distances. Here's a detailed, step-by-step explanation.
### Step 1: Initial Potential Energies
1. Gravitational Potential Energy (GPE)
- Formula: \( GPE = m \cdot g \cdot h \)
- Given:
- \( m = 1 \, \text{kg} \) (mass of the object)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 0.5 \, \text{m} \) (initial height above the ground)
- Calculation:
[tex]\[ GPE_{\text{initial}} = 1 \cdot 9.8 \cdot 0.5 = 4.9 \, \text{Joules} \][/tex]
2. Elastic Potential Energy (EPE)
- Formula: \( EPE = \frac{1}{2} k x^2 \)
- Given:
- \( k = 1 \, \text{N/m} \) (spring constant)
- \( x = 0.5 \, \text{m} \) (initial stretch of the spring)
- Calculation:
[tex]\[ EPE_{\text{initial}} = \frac{1}{2} \cdot 1 \cdot (0.5)^2 = 0.125 \, \text{Joules} \][/tex]
At the initial state, we are told that the potential energies are equal.
### Step 2: Doubling the Distances
1. New Gravitational Potential Energy (GPE)
- New height: \( h_{\text{new}} = 2 \times 0.5 = 1 \, \text{m} \)
- Calculation:
[tex]\[ GPE_{\text{new}} = 1 \cdot 9.8 \cdot 1 = 9.8 \, \text{Joules} \][/tex]
2. New Elastic Potential Energy (EPE)
- New stretch: \( x_{\text{new}} = 2 \times 0.5 = 1 \, \text{m} \)
- Calculation:
[tex]\[ EPE_{\text{new}} = \frac{1}{2} \cdot 1 \cdot (1)^2 = 0.5 \, \text{Joules} \][/tex]
### Step 3: Compare the Ratios of the new Potential Energies
1. Ratio of the new GPE to the initial GPE:
[tex]\[ \frac{GPE_{\text{new}}}{GPE_{\text{initial}}} = \frac{9.8}{4.9} = 2 \][/tex]
2. Ratio of the new EPE to the initial EPE:
[tex]\[ \frac{EPE_{\text{new}}}{EPE_{\text{initial}}} = \frac{0.5}{0.125} = 4 \][/tex]
### Step 4: Analysis of Potential Energy Comparison
After doubling the distances, the gravitational potential energy of the object becomes twice its original value, while the elastic potential energy of the spring becomes four times its original value.
Thus, the elastic potential energy of the spring will be four times greater than the gravitational potential energy of the object.
So, the correct answer is:
- The elastic potential energy of the spring will be four times greater than the gravitational potential energy of the object.
### Step 1: Initial Potential Energies
1. Gravitational Potential Energy (GPE)
- Formula: \( GPE = m \cdot g \cdot h \)
- Given:
- \( m = 1 \, \text{kg} \) (mass of the object)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 0.5 \, \text{m} \) (initial height above the ground)
- Calculation:
[tex]\[ GPE_{\text{initial}} = 1 \cdot 9.8 \cdot 0.5 = 4.9 \, \text{Joules} \][/tex]
2. Elastic Potential Energy (EPE)
- Formula: \( EPE = \frac{1}{2} k x^2 \)
- Given:
- \( k = 1 \, \text{N/m} \) (spring constant)
- \( x = 0.5 \, \text{m} \) (initial stretch of the spring)
- Calculation:
[tex]\[ EPE_{\text{initial}} = \frac{1}{2} \cdot 1 \cdot (0.5)^2 = 0.125 \, \text{Joules} \][/tex]
At the initial state, we are told that the potential energies are equal.
### Step 2: Doubling the Distances
1. New Gravitational Potential Energy (GPE)
- New height: \( h_{\text{new}} = 2 \times 0.5 = 1 \, \text{m} \)
- Calculation:
[tex]\[ GPE_{\text{new}} = 1 \cdot 9.8 \cdot 1 = 9.8 \, \text{Joules} \][/tex]
2. New Elastic Potential Energy (EPE)
- New stretch: \( x_{\text{new}} = 2 \times 0.5 = 1 \, \text{m} \)
- Calculation:
[tex]\[ EPE_{\text{new}} = \frac{1}{2} \cdot 1 \cdot (1)^2 = 0.5 \, \text{Joules} \][/tex]
### Step 3: Compare the Ratios of the new Potential Energies
1. Ratio of the new GPE to the initial GPE:
[tex]\[ \frac{GPE_{\text{new}}}{GPE_{\text{initial}}} = \frac{9.8}{4.9} = 2 \][/tex]
2. Ratio of the new EPE to the initial EPE:
[tex]\[ \frac{EPE_{\text{new}}}{EPE_{\text{initial}}} = \frac{0.5}{0.125} = 4 \][/tex]
### Step 4: Analysis of Potential Energy Comparison
After doubling the distances, the gravitational potential energy of the object becomes twice its original value, while the elastic potential energy of the spring becomes four times its original value.
Thus, the elastic potential energy of the spring will be four times greater than the gravitational potential energy of the object.
So, the correct answer is:
- The elastic potential energy of the spring will be four times greater than the gravitational potential energy of the object.
