Question

A mass $m$ slides from rest at height $h$ down a smooth curved surface which becomes horizontal at zero height. A spring is fixed horizontally on the level part of the surface. The spring constant is $kN/m$. When the mass encounters the spring it compresses it by an amount $x=h/10$.
If $m=1kg,h=5m$ then find $k/100$.

Answer 1

R.E.F image

$x=\frac{h}{10}=\frac{5}{10}=0.5m$

kinetic energy delivered to spring

$\Rightarrow \frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2$ but $v=\sqrt{29h}$

So. $k{x}^2=m\left(29h\right)$

$\Rightarrow k=\frac{2mgh}{x^2}=\frac{2\times 1\times 10\times 5}{(0.5{)}^2}$

$\Rightarrow k=4w\Rightarrow \frac{k}{100}=4$