Question

simple harmonic motion a spring that has a spring constant of 440 n/m exerts a force of 88 n on a box. what is the displacement of the spring? 5 m 528 m 0.2 m 352 m

Explanation:

Step1: Recall Hooke's Law

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

Step2: Substitute the given values

We know that \( F = 88\space N \) and \( k = 440\space N/m \). Substituting these values into the formula \( x=\frac{F}{k} \), we get \( x=\frac{88}{440} \).

Step3: Calculate the displacement

Simplifying \( \frac{88}{440} \), we divide both the numerator and the denominator by 88. \( \frac{88\div88}{440\div88}=\frac{1}{5} = 0.2\space m \).

Answer:

\( 0.2\space m \) (corresponding to the option with \( 0.2\space m \))