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A force of 0.8n stretches an elastic spring by 20m. Find the elastic constant of the springAsked by Aisha
A force 0.8₦ stretches an elastics spring by3m find the elastic constant of the spring
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Answered by
GPT-4o mini
To find the elastic constant (spring constant) of a spring, you can use Hooke's Law, which states:
\[ F = k \cdot x \]
Where:
- \( F \) is the force applied to the spring (in newtons),
- \( k \) is the spring constant (in newtons per meter),
- \( x \) is the extension (or compression) of the spring (in meters).
In this case, you have:
- \( F = 0.8 \, \text{N} \)
- \( x = 3 \, \text{m} \)
Rearranging the formula to find \( k \):
\[ k = \frac{F}{x} \]
Substituting the values:
\[ k = \frac{0.8 \, \text{N}}{3 \, \text{m}} \]
\[ k \approx 0.2667 \, \text{N/m} \]
Therefore, the elastic constant (spring constant) of the spring is approximately **0.267 N/m**.
\[ F = k \cdot x \]
Where:
- \( F \) is the force applied to the spring (in newtons),
- \( k \) is the spring constant (in newtons per meter),
- \( x \) is the extension (or compression) of the spring (in meters).
In this case, you have:
- \( F = 0.8 \, \text{N} \)
- \( x = 3 \, \text{m} \)
Rearranging the formula to find \( k \):
\[ k = \frac{F}{x} \]
Substituting the values:
\[ k = \frac{0.8 \, \text{N}}{3 \, \text{m}} \]
\[ k \approx 0.2667 \, \text{N/m} \]
Therefore, the elastic constant (spring constant) of the spring is approximately **0.267 N/m**.
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