question_answer

An object of mass \[m\]falls on to a spring of constant k from height\[h.\]The spring undergoes compression by a length\[x.\]The maximum compression \[x\] is given by the equation:

A) \[m\,g\,h=\frac{1}{2}k{{x}^{2}}\]

B)  \[m\,g(h+x)=\frac{1}{2}k{{x}^{2}}\]

C)  \[m\,g(h+x)=-kx\]

D)  \[m\,g\,h=-kx\]

Correct Answer: B

Solution :

 Key Idea: Energy is conserved. From law of conservation of energy potential energy at height \[h=\]energy stored in compressed spring. \[\therefore \] \[mgh+mgx=\frac{1}{2}k{{x}^{2}}\] \[\Rightarrow \] \[mg(h+x)=\frac{1}{2}k{{x}^{2}}\]