Question

1. a 40 n force is applied to a spring with spring constant 80 n/m. how much will the spring stretch?

Explanation:

Step1: Recall Hooke's Law

Hooke's Law states that the force \( F \) applied to a spring is related to the spring constant \( k \) and the displacement \( x \) (stretch or compression) by the formula \( F = kx \). We need to solve for \( x \), so we can rearrange the formula to \( x=\frac{F}{k} \).

Step2: Substitute the given values

We are given that \( F = 40\space N \) and \( k = 80\space N/m \). Substituting these values into the formula \( x=\frac{F}{k} \), we get \( x=\frac{40}{80} \).

Step3: Calculate the result

Simplifying \( \frac{40}{80} \), we find that \( x = 0.5\space m \).

Answer:

The spring will stretch \( 0.5\space m \) (or 50 centimeters).