Potential Energy & Equilibrium: Practice Problem & Solution
Given below are two statements: Assertion (A): A block of mass \(m\) is dropped from a height \(h\) on to a spring of spring constant \(k\). The maximum compression \(x_{\text{max}}\) in the spring is given by \(x_{\text{max}} = \sqrt{\frac{2mgh}{k}}\) Reason (R): The work done by gravitational force is equal to the elastic potential energy stored in the spring at maximum compression. In the light of the above statements, choose the correct answer from the options given below.
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:
By conservation of energy, the total loss in gravitational potential energy is \(mg(h + x_{\text{max}})\), which equals the elastic potential energy stored in the spring \(\frac{1}{2}kx_{\text{max}}^2\). Thus, the given formula for \(x_{\text{max}}\) is incorrect as it neglects the term \(mgx_{\text{max}}\). Hence, (A) is false but (R) is true.
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