To find the weight required to extend the spring by 2.5 cm, we can first determine the spring constant \( k \) using Hooke’s law, which states that the force applied to a spring is proportional to its extension. Hooke's law is given by:
\[ F = k \cdot x \]
where:
- \( F \) is the force (weight) applied,
- \( k \) is the spring constant,
- \( x \) is the extension (displacement from the equilibrium position).
First, we determine the force applied by the 40 grams mass.
- Convert mass to weight (force):
\[ \text{Mass} = 40 \text{ grams} = 0.04 \text{ kg} \] \[ F = m \cdot g = 0.04 \text{ kg} \cdot 9.81 \text{ m/s}^2 \approx 0.3924 \text{ N} \]
- Calculate spring constant \( k \) with the original extension of 2 cm (0.02 m):
\[ x_1 = 0.02 \text{ m} \] \[ k = \frac{F}{x_1} = \frac{0.3924 \text{ N}}{0.02 \text{ m}} = 19.62 \text{ N/m} \]
- Determine the force required to extend it by 2.5 cm (0.025 m) using the spring constant \( k \):
\[ x_2 = 0.025 \text{ m} \] \[ F_2 = k \cdot x_2 = 19.62 \text{ N/m} \cdot 0.025 \text{ m} = 0.4905 \text{ N} \]
- Convert this force back to mass:
\[ \text{Weight} = F_2 = 0.4905 \text{ N} \] \[ \text{Mass} = \frac{F_2}{g} = \frac{0.4905 \text{ N}}{9.81 \text{ m/s}^2} \approx 0.05 \text{ kg} \approx 50 \text{ grams} \]
Therefore, the weight required to extend the spring by 2.5 cm is approximately 50 grams.
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