Question

When a load of $10kg$ is suspended on a metallic wire, its length increase by $2mm$. The force constant of the wire is

Answer 1

using Hooke'n low, F = Kx, where F = mg weight of load,a = elongation

$\Rightarrow k=\frac{mg}{x}=\frac{10\times 10}{2\times {10}^3}=5\times {10}^{6}n/m$