question_answer

An ideal spring with spring-constant K is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is [IIT-JEE (Screening) 2002]

A)            4 Mg/K                                   

B)            2 Mg/K

C)            Mg/K                                      

D)            Mg/2K

Correct Answer: B

Solution :

                   Let x be the maximum extension of the spring. From energy conservation Loss in gravitational potential energy = Gain in potential energy of spring \[Mgx=\frac{1}{2}K{{x}^{2}}\] \[\Rightarrow x=\frac{2Mg}{K}\]