A spring of force constant \(K=1000\,N/m\) is compressed by \(5\,cm\). If a mass of \(2.5\,kg\) is attached and released, find its speed as it passes the equilibrium position.
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For a spring-mass system, remember:
\[
\boxed{\frac12Kx^2=\frac12mv^2}
\]
or directly,
\[
\boxed{v=x\sqrt{\frac{K}{m}}}
\]
At the equilibrium position, the spring has zero potential energy and the kinetic energy of the mass is maximum.
At maximum height the elastic PE of the spring converts entirely to gravitational PE: $\frac{1}{2}Kx^2 = mgh$. Substituting gives $h = Kx^2/(2mg) = 1000 \times (0.05)^2/(2 \times 2.5 \times 10) = 0.05$ m $= 5$ cm.