question_answer

Find the maximum compression in the spring, if the lower block is shifted to rightwards with Acceleration \['a'\] All the surfaces are smooth:

A) \[\frac{ma}{2k}\]                     

B) \[\frac{2ma}{k}\]

C) \[\frac{ma}{k}\]                       

D) \[\frac{4ma}{k}\]

Correct Answer: B

Solution :

[b] \[mv\frac{dv}{dx}=(ma-kx)\] \[\int\limits_{0}^{0}{mvdv=\int\limits_{0}^{x}{(mx-kx)dx}}\] \[0=\max -\frac{k{{x}^{2}}}{2}\Rightarrow x=\frac{2ma}{k}\]