Directions: In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
A block is suspended by an ideal spring of force constant k. If the block is pulled down by applying a constant force F and if maximum displacement of block from its initial position of rest is\[\delta \]then: (1) \[\frac{F}{k}<\delta <\frac{2F}{k}\] (2) \[\delta =\frac{2F}{k}\] (3) work done by the force is equal to \[F\delta \] (4) increases in energy stored in spring is \[\frac{1}{2}k{{\delta }^{2}}\]A) 1 and 2 are correct
B) 2 and 3 are correct
C) 1 and 4 and correct
D) 1, 2 and 4 are correct
Correct Answer: B
Solution :
Initial elongation of the spring\[=\frac{mg}{k}\]. When the force is applied to pull the block down, the work done by\[F+\]lose of gravitational PE = increase in strain energy of the spring So, \[F\delta +mg\,\delta =\frac{1K}{2}{{\left( \frac{mg}{k}+\delta \right)}^{2}}-\frac{{{m}^{2}}{{g}^{2}}}{2k}\] Solving we get, \[\delta =\frac{2F}{k}\] Also work done by pulling force \[F=F\delta \] Increase in strain energy of spring \[=\frac{1}{2}k{{\left( \frac{mg}{k}+\delta \right)}^{2}}-\frac{{{m}^{2}}{{g}^{2}}}{2k}\ne \frac{1}{2}k{{\delta }^{2}}\]
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