Question

3 formula 1 point a spring has some spring constant \k\. a mass of m kilograms is hung from the spring and allowed to come to rest and the spring stretches some distance \d\ meters. if the mass is increased to 3m kilograms, how many times greater must the total stretched distance be? answer previous next

Explanation:

Step1: Apply Hooke's Law

Hooke's Law is \(F = kd\), where \(F\) is the force, \(k\) is the spring - constant, and \(d\) is the displacement. When the mass is \(m\), the force \(F_1=mg\) (where \(g\) is the acceleration due to gravity). So, \(mg = kd_1\), and \(d_1=\frac{mg}{k}\).

Step2: Calculate for new mass

When the mass is \(3m\), the force \(F_2 = 3mg\). Using Hooke's Law \(F_2=kd_2\), so \(3mg=kd_2\), and \(d_2=\frac{3mg}{k}\).

Step3: Find the ratio of displacements

\(\frac{d_2}{d_1}=\frac{\frac{3mg}{k}}{\frac{mg}{k}}\). The \(\frac{mg}{k}\) terms cancel out, and \(\frac{d_2}{d_1}=3\).

Answer:

\(3\)