Work Energy And Power Question 145

Question: Find the maximum compression in the spring, if the lower block is shifted to rightwards with Acceleration $ ‘a’ $ All the surfaces are smooth:

Options:

A) $ \frac{ma}{2k} $

B) $ \frac{2ma}{k} $

C) $ \frac{ma}{k} $

D) $ \frac{4ma}{k} $

Show Answer

Answer:

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{kx^{2}}{2}\Rightarrow x=\frac{2ma}{k} $



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