Question

a force of 75 n is applied to a spring, causing it to stretch 0.3 m. what is the spring constant of the spring? 22.5 n/m 0.004 n/m 250 n/m 75.3 n/m

Explanation:

Step1: Recall Hooke's Law

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

Step2: Substitute the given values

We are given \( F = 75\space N \) and \( x = 0.3\space m \). Substituting these values into the formula for \( k \), we get \( k=\frac{75}{0.3} \).

Step3: Calculate the value of \( k \)

Calculating \( \frac{75}{0.3} \), we can multiply the numerator and denominator by 10 to eliminate the decimal: \( \frac{75\times10}{0.3\times10}=\frac{750}{3} = 250 \space N/m \).

Answer:

250 N/m (corresponding to the option with "250 N/m")