Potential Energy & Equilibrium: Practice Problem & Solution
A block of mass \(M\) is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant \(k\). The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be: (2009)
Solution Explained:
To solve this problem, we apply the core principles of Potential Energy & Equilibrium. Understanding the underlying formula is key to arriving at the correct answer below:
Concept: Conservation of Mechanical Energy. Initial state: \(KE_i = 0\), \(PE_i = 0\) (reference at initial position). Final state (maximum extension \(x_{max}\)): \(KE_f = 0\), \(PE_f = -Mgx_{max} + frac{1}{2}kx_{max}^2\). By energy conservation, \(KE_i + PE_i = KE_f + PE_f\), so \(0 = -Mgx_{max} + \frac{1}{2}kx_{max}^2\). Solving for \(x_{max}\), we get \(Mg = \frac{1}{2}kx_{max}\), which yields \(x_{max} =\frac{2Mg}{k}\).
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