Answer:
To find the unstretched length of the spring, we can use the information given about the forces and the extensions of the spring.
-
Initial Information:
- When a force of 4 N is applied, the length of the spring is 14 cm.
- When a force of 6 N is applied, the spring extends by 4 cm.
-
Determine the extension for each force:
- Let ( L_0 ) be the unstretched length of the spring.
- When 4 N is applied, the length of the spring is 14 cm, so the extension ( x_1 ) can be expressed as:
[
x_1 = 14 \text{ cm} - L_0
]
- When 6 N is applied, the spring extends by 4 cm, so the new length is:
[
L_2 = L_0 + 4 \text{ cm}
]
-
Set up the equations:
- For the 4 N force:
[
14 \text{ cm} = L_0 + x_1
]
- For the 6 N force:
[
L_0 + 4 \text{ cm} = L_0 + x_2
]
- The extension ( x_2 ) when 6 N is applied can be expressed as:
[
x_2 = 4 \text{ cm}
]
-
Using Hooke's Law:
- According to Hooke's Law, the extension is proportional to the force applied:
[
\frac{x_2 - x_1}{6 \text{ N} - 4 \text{ N}} = \frac{4 \text{ cm} - x_1}{4 \text{ N}}
]
- Rearranging gives:
[
\frac{4 \text{ cm} - x_1}{2 \text{ N}} = \frac{4 \text{ cm} - x_1}{4 \text{ N}}
]
-
Solving for ( x_1 ):
- From the first equation:
[
14 \text{ cm} = L_0 + x_1 \implies x_1 = 14 \text{ cm} - L_0
]
- Substitute ( x_1 ) into the second equation:
[
4 \text{ cm} - (14 \text{ cm} - L_0) = 2 \text{ N} \cdot \frac{4 \text{ cm} - (14 \text{ cm} - L_0)}{4 \text{ N}}
]
- Simplifying gives:
[
4 \text{ cm} - 14 \text{ cm} + L_0 = 2 \text{ N} \cdot \frac{L_0 - 10 \text{ cm}}{4 \text{ N}}
]
- This leads to:
[
L_0 - 10 \text{ cm} = 2 \cdot \frac{L_0 - 10 \text{ cm}}{4}
]
- Solving this gives:
[
L_0 - 10 \text{ cm} = \frac{L_0 - 10 \text{ cm}}{2}
]
- Thus, ( L_0 = 10 \text{ cm} ).
-
Final Calculation:
- The unstretched length of the spring is:
[
L_0 = 10 \text{ cm}
]
Therefore, the unstretched length of the spring is 10 cm.spring, we can use the information given about the forces and the extensions of the spring.
-
Initial Information:
- When a force of 4 N is applied, the length of the spring is 14 cm.
- When a force of 6 N is applied, the spring extends by 4 cm.
-
Determine the extension for each force:
- Let ( L_0 ) be the unstretched length of the spring.
- When 4 N is applied, the spring's length is 14 cm, so the extension ( x_1 ) can be expressed as:
[
x_1 = 14 \text{ cm} - L_0
]
- When 6 N is applied, the spring extends by 4 cm, so the new length is:
[
L_2 = L_0 + 4 \text{ cm}
]
-
Set up the equations:
- For the 4 N force:
[
14 \text{ cm} = L_0 + x_1
]
- For the 6 N force:
[
L_0 + 4 \text{ cm} = L_0 + x_2
]
- The extension ( x_2 ) when 6 N is applied can be expressed as:
[
x_2 = 4 \text{ cm}
]
-
Using Hooke's Law:
- Hooke's Law states that the force ( F ) is proportional to the extension ( x ):
[
F = kx
]
- For the 4 N force:
[
4 = kx_1
]
- For the 6 N force:
[
6 = kx_2
]
-
Relate the extensions:
- From the second equation, we know ( x_2 = 4 \text{ cm} ):
[
6 = k \cdot 4 \implies k = \frac{6}{4} = 1.5 \text{ N/cm}
]
- Substitute ( k ) back into the first equation:
[
4 = 1.5x_1 \implies x_1 = \frac{4}{1.5} \approx 2.67 \text{ cm}
]
-
Substituting back to find ( L_0 ):
- Now we can substitute ( x_1 ) back into the equation for the 4 N force:
[
14 \text{ cm} = L_0 + 2.67 \text{ cm}
]
- Rearranging gives:
[
L_0 = 14 \text{ cm} - 2.67 \text{ cm} \approx 11.33 \text{ cm}
]
Thus, the unstretched length of the spring is approximately 11.33 cm.