Question

a force of 20 n causes a spring to stretch 5 cm. what is the spring constant? 20 n/m 2 n/m 400 n/m 1 n/m

Explanation:

Step1: Recall Hooke's Law

Hooke's Law states that the force \( F \) applied to a spring is proportional to the displacement \( x \) from its equilibrium position, given by the formula \( F = kx \), where \( k \) is the spring constant. We need to solve for \( k \), so we can rearrange the formula to \( k=\frac{F}{x} \).

Step2: Convert units

The displacement \( x \) is given as 5 cm. We need to convert this to meters because the unit of the spring constant is N/m. Since 1 m = 100 cm, \( 5\ \text{cm}=\frac{5}{100}\ \text{m} = 0.05\ \text{m} \).

Step3: Substitute values into the formula

We know that \( F = 20\ \text{N} \) and \( x = 0.05\ \text{m} \). Substituting these into the formula \( k=\frac{F}{x} \), we get \( k=\frac{20\ \text{N}}{0.05\ \text{m}} \).

Step4: Calculate the spring constant

Calculating \( \frac{20}{0.05} \), we have \( 20\div0.05 = 400 \). So the spring constant \( k = 400\ \text{N/m} \).

Answer:

400 N/m (corresponding to the option "400 N/m")