Question
2) A Hooke's spring has a spring constant of \( 40.0 \mathrm{Nm}^{-1} \) is suspended to a fixed point. (a) Calculate the extension of the spring when a 55.0 N weight is suspended from the free end.
Ask by Norris Tyler. in Trinidad and Tobago
Nov 13,2024
Real Tutor Solution
Tutor-Verified Answer
Answer
The extension of the spring is \( 1.375 \, \text{m} \).
Solution
To calculate the extension of a Hooke's spring when a weight is suspended from it, we can use Hooke's Law, which states:
\[
F = k \cdot x
\]
where:
- \( F \) is the force applied (in Newtons),
- \( k \) is the spring constant (in N/m),
- \( x \) is the extension of the spring (in meters).
Given:
- \( F = 55.0 \, \text{N} \)
- \( k = 40.0 \, \text{N/m} \)
We can rearrange the formula to solve for \( x \):
\[
x = \frac{F}{k}
\]
Substituting the values:
\[
x = \frac{55.0 \, \text{N}}{40.0 \, \text{N/m}} = 1.375 \, \text{m}
\]
Thus, the extension of the spring when a 55.0 N weight is suspended from the free end is \( 1.375 \, \text{m} \).
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Simplify this solution Bonus Knowledge
To find the extension of the spring, we can use Hooke's Law, which states that the force (F) applied to a spring is directly proportional to the extension (x) of the spring. The formula is given by: \[ F = kx \] where \( k \) is the spring constant and \( x \) is the extension. Rearranging this gives us: \[ x = \frac{F}{k} \] Given that the weight \( F = 55.0 \, \text{N} \) and the spring constant \( k = 40.0 \, \text{Nm}^{-1} \): \[ x = \frac{55.0 \, \text{N}}{40.0 \, \text{Nm}^{-1}} = 1.375 \, \text{m} \] So the extension of the spring is \( 1.375 \, \text{m} \).
