Answer to Question #179986 in Physics for Richard

Question

A vertically placed spring whose constant is 92 N / m is attached to the base at the lower end. A 0.6 kg wooden block is placed on top of the spring. If the system shown is subject to simple harmonic oscillations, determine at what amplitude the contact between the spring and the block will be lost. (Note: The block at the top of the spring is not attached, but only placed!) (Answer: x_o= 0.0639 m)

Answer 1

$k(x+x_0)^2/2=mg(x+x_0)$

$F=kx_0\to mg=kx_0\to x_0=mg/k=0.6\cdot9.81/92=0.063978\ (m)$

So, we have

$k(x+0.063978)^2/2=mg(x+0.063978)\to$

$92\cdot(x+0.063978)^2/2=0.6\cdot9.81\cdot (x+0.063978)\to$

$x=0.063978\ (m)$ . Answer