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} $
