A spiral spring of length [tex]$30 \text{ cm}[tex]$[/tex] has a scale pan hanging freely at its lower end. When an object of mass [tex]$[/tex]60 \text{ g}[tex]$[/tex] is placed in the pan, its length becomes [tex]$[/tex]32.5 \text{ cm}[tex]$[/tex]. When another object of mass [tex]$[/tex]100 \text{ g}[tex]$[/tex] is placed in the pan, the length becomes [tex]$[/tex]34.5 \text{ cm}$[/tex].

Calculate the force constant of the spring.



Answer :

Certainly! Let's walk through the steps to solve for the force constant (spring constant) of the spring and find the extensions caused by the given masses.

### Step-by-Step Solution:

1. Initial Length of the Spring:
The initial length of the spring, \( L_0 \), is \( 30 \text{ cm} \).

2. Final Length with 60g Mass:
When an object of mass \( 60 \text{ g} \) is added, the length becomes \( 32.5 \text{ cm} \).

3. Final Length with 100g Mass:
When an object of mass \( 100 \text{ g} \) is added, the length becomes \( 34.5 \text{ cm} \).

4. Calculate Extensions:
- Extension with 60g:
[tex]\[ \Delta x_{60} = 32.5 \text{ cm} - 30 \text{ cm} = 2.5 \text{ cm} \][/tex]
Convert to meters:
[tex]\[ \Delta x_{60} = 2.5 \text{ cm} \times 0.01 = 0.025 \text{ m} \][/tex]

- Extension with 100g:
[tex]\[ \Delta x_{100} = 34.5 \text{ cm} - 30 \text{ cm} = 4.5 \text{ cm} \][/tex]
Convert to meters:
[tex]\[ \Delta x_{100} = 4.5 \text{ cm} \times 0.01 = 0.045 \text{ m} \][/tex]

5. Convert Mass to Kilograms:
- Mass of 60g:
[tex]\[ m_{60} = 60 \text{ g} \times 0.001 = 0.06 \text{ kg} \][/tex]

- Mass of 100g:
[tex]\[ m_{100} = 100 \text{ g} \times 0.001 = 0.1 \text{ kg} \][/tex]

6. Calculate the Forces:
Using \( F = m \times g \) where \( g = 9.8 \text{ m/s}^2 \):
- Force with 60g:
[tex]\[ F_{60} = 0.06 \text{ kg} \times 9.8 \text{ m/s}^2 = 0.588 \text{ N} \][/tex]

- Force with 100g:
[tex]\[ F_{100} = 0.1 \text{ kg} \times 9.8 \text{ m/s}^2 = 0.98 \text{ N} \][/tex]

7. Calculate the Force Constant (k):
Using Hooke's Law \( F = k \times \Delta x \):
- For 60g:
[tex]\[ k_{60} = \frac{F_{60}}{\Delta x_{60}} = \frac{0.588 \text{ N}}{0.025 \text{ m}} = 23.52 \text{ N/m} \][/tex]

- For 100g:
[tex]\[ k_{100} = \frac{F_{100}}{\Delta x_{100}} = \frac{0.98 \text{ N}}{0.045 \text{ m}} = 21.78 \text{ N/m} \][/tex]

8. Average Force Constant:
Since the spring constant \( k \) should be consistent, we average the two calculated values to get a more accurate result:
[tex]\[ k = \frac{k_{60} + k_{100}}{2} = \frac{23.52 \text{ N/m} + 21.78 \text{ N/m}}{2} = 22.65 \text{ N/m} \][/tex]

### Summary:
- Extension with 60g mass: \( 0.025 \text{ m} \)
- Extension with 100g mass: \( 0.045 \text{ m} \)
- Force with 60g mass: \( 0.588 \text{ N} \)
- Force with 100g mass: \( 0.98 \text{ N} \)
- Average force constant: \( 22.65 \text{ N/m} \)

These steps provide detailed calculations leading to the determination of the spring's force constant.