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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 isA. Mg/2kB. Mg/kC. 2 Mg/kD. 4 Mg/k |
Answer» Correct Answer - C Loss in grav. PE= gain in spring PE At maximum elongation `Mgx=(1)/(2)kx^(2)` `x=(2Mg)/(k)` |
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