• # question_answer         When a weight of 100 N falls on a spring of stiffness 1 kN/m from a height of 2m, the deflection caused in the first fall is:                              A) Equal to 0.1 m  B) Between 0.1 and 0.2C) Equal to 0.2 m     D) More than 0.2 m

$W(h+\delta )=\frac{1}{2}k{{\delta }^{2}}$ $100(2+\delta )=\frac{1}{2}\times 1000\times {{\delta }^{2}}$ $=5{{\delta }^{2}}-\delta -2=0$ $\delta =\frac{1\pm \sqrt{1+40}}{10}=0.74\,m>0.2\,m$