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I need to calculate the velocity of an object when it is thrown by a spring; we have to calculate the velocity from $U=\frac 12 kx^2$. Now I know that $U=L=F\cdot S$ and $S=\frac{v_f^2-v_0^2}{2a}$. Here's what I did: $$U=L=F\cdot S=ma\cdot S\rightarrow L=ma\cdot \frac{v_f^2}{2a}=\frac{m\cdot v_f^2}{2}$$

$$v_f=\sqrt{\frac{2L}{m}}\rightarrow \sqrt{\frac{2\frac 12 kx^2}{m}}=x\cdot\sqrt{\frac km} $$

Is this correct?

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closed as off-topic by ACuriousMind, Kyle Kanos, David Z Oct 14 '14 at 18:19

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Yes, alternatively you can use that the potential energy of the spring is transformed into kinetic energy of the object. This is simpler than considering the work $L$ done by the spring. The result is however the same: $$ U=E_{kin}\\ \Rightarrow \frac{1}{2} k x^2=\frac{1}{2}m v^2\\ \Rightarrow v=x\sqrt\frac{k}{m} $$

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  • $\begingroup$ thx much easier $\endgroup$ – Peterix Oct 14 '14 at 17:18

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