Question

A non-Hookian spring has force $F=-k{x}^2$, where $k$ is the spring constant and $x$ is the displacement from its unstretched position. For the system shown, a mass $m$ connected to an unstretched spring is initially at rest. If the distance up to which the spring extends before the system momentarily comes to rest is ${\left(\frac{pmg}{qk}\right)}^{\frac{1}{2}}$, then find $(p+q)$. Assume that all surfaces as well as the pulley are frictionless.

Answer 1

$\Delta W=-\Delta U$

$mgx\cos {60}^{\circ }={\int }_0^{x}k{x}^{2}dx$

$\frac{mgx}{2}=\left(\frac{k{x}^3}{3}\right)$

$x={\left(\frac{3mg}{2k}\right)}^{\frac{1}{2}}$