Question

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

Answer 1

$ U _{ f }=\frac{1}{2} k \left(\frac{ x _0}{2}\right)^2 $
$ K _{ f }=0.25\, J $
$ \frac{1}{2} kx _0^2+0=\frac{1}{2} k \frac{ x _0^2}{4}+0.25 $
$\frac{1}{2} kx _0^2 \frac{3}{4}=\frac{1}{4} $
$ \frac{1}{2} k \frac{3}{100}=1 \Rightarrow k =\frac{200}{3} N / m $
$ =67 \,N / m $