Question

a spring has a spring constant of 83 (i intentionally left out the units of the spring constant here because there is a problem in the exit slip that asks the units). how far must the spring be compressed or stretched from its equilibrium position for the spring to exert a force with a magnitude of 13 newtons? report your answer as a positive number.

Explanation:

Step1: Recall Hooke's Law

Hooke's Law is \( F = kx \), where \( F \) is the force exerted by the spring, \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position. We need to solve for \( x \), so we can rewrite the formula as \( x=\frac{F}{k} \).

Step2: Substitute the given values

We are given that \( F = 13\) N and \( k = 83\). Substituting these values into the formula \( x=\frac{F}{k}\), we get \( x=\frac{13}{83}\).

Step3: Calculate the value of \( x \)

\( x=\frac{13}{83}\approx 0.157 \)

Answer:

\(0.157\)