Question

A block is fastened to a horizontal spring. The block is pulled to a distance $x=10cm$ from its equilibrium position (at $x=0$ ) on a frictionless surface from rest. The energy of the block at $x=5$ $cm$ is $0.25J$. The spring constant of the spring is ___$N{m}^{-1}$.

Answer 1

Answer: 50

$U_f=\frac{1}{2}k{\left(\frac{x_0}{2}\right)}^2$
$K_f=0.25J$
$\frac{1}{2}k{x}_0^2+0=\frac{1}{2}k\frac{x_0^2}{4}+0.25$
$\frac{1}{2}k{x}_0^2\frac{3}{4}=\frac{1}{4}$
$\frac{1}{2}k\frac{3}{100}=1\Rightarrow k=\frac{200}{3}N/m$
$=67N/m$