Question

1. a 3.00-newton force causes a spring to stretch 60.0 centimeters. calculate the spring constant of this spring. show all work, including the equation and substitution with units.

Explanation:

Step1: Recall Hooke's Law

Hooke's Law is given by \( F = k\Delta x \), where \( F \) is the force applied, \( k \) is the spring constant, and \( \Delta x \) is the displacement (stretch or compression) of the spring. We need to solve for \( k \), so we rearrange the formula to \( k=\frac{F}{\Delta x} \).

Step2: Convert units

The displacement \( \Delta x \) is given as 60.0 centimeters. We need to convert this to meters because the SI unit of force is newtons (kg·m/s²) and the SI unit of length is meters. Since \( 1\space m = 100\space cm \), we have \( \Delta x=\frac{60.0\space cm}{100\space cm/m}=0.600\space m \).

Step3: Substitute values into the formula

We know that \( F = 3.00\space N \) and \( \Delta x=0.600\space m \). Substituting these into the formula \( k = \frac{F}{\Delta x} \), we get \( k=\frac{3.00\space N}{0.600\space m} \).

Step4: Calculate the spring constant

Performing the division, \( \frac{3.00}{0.600}=5.00 \). So the spring constant \( k = 5.00\space N/m \).

Answer:

The spring constant of the spring is \( \boldsymbol{5.00\space N/m} \).