A) \[\frac{ma}{2k}\]
B) \[\frac{2ma}{k}\]
C) \[\frac{ma}{k}\]
D) \[\frac{4ma}{k}\]
Correct Answer: B
Solution :
[b] \[mv\frac{dv}{dx}=(ma-kx)\] \[\int\limits_{0}^{0}{mvdv=\int\limits_{0}^{x}{(mx-kx)dx}}\] \[0=\max -\frac{k{{x}^{2}}}{2}\Rightarrow x=\frac{2ma}{k}\]You need to login to perform this action.
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