To determine the length of the spring when a force of 4.0 N is applied, we can use Hooke's Law, which states that the extension of a spring is directly proportional to the force applied to it, within the elastic limit of the material. Hooke's Law can be expressed as:
\[ F = k \cdot x \]
where:
- \( F \) is the force applied,
- \( k \) is the spring constant,
- \( x \) is the extension of the spring from its natural length.
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First, we need to calculate the spring constant \( k \) using the initial force and extension:
Given:
- Natural length of the spring = 1.5 cm
- Extension at 1.5N = 0.02 cm
Extension in meters: \[ x_1 = 0.02 , \text{cm} = 0.0002 , \text{m} \]
Using Hooke's Law: \[ F_1 = k \cdot x_1 \] \[ 1.5 = k \cdot 0.0002 \]
Solving for \( k \): \[ k = \frac{1.5}{0.0002} = 7500 , \text{N/m} \]
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Now, we can find the extension when a force of 4.0 N is applied:
Given:
- \( F_2 = 4.0 , \text{N} \)
Using Hooke's Law again: \[ F_2 = k \cdot x_2 \] \[ 4.0 = 7500 \cdot x_2 \]
Solving for \( x_2 \): \[ x_2 = \frac{4.0}{7500} = 0.00053333 , \text{m} \] \[ x_2 \approx 0.0533 , \text{cm} \]
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Finally, we can find the new length of the spring when the force of 4.0 N is applied:
New length \( L \) is the natural length plus the extension: \[ L = 1.5 , \text{cm} + 0.0533 , \text{cm} \] \[ L \approx 1.5533 , \text{cm} \]
So the length of the spring when a force of 4.0 N is applied will be approximately 1.553 cm.