If a spring has a stiffness of 1000 Nm–1. What work will be done by extending the spring by 100mm? – NABTEB QUESTION

Work done = ½ FX = ½ KX²

K = stiffness constant = 100Nm-1
X = Extension of spring = 100mm = 0.1m
Work done = ½ x 1000 x (.1)² Joules

= ½ x 1000 .01J
W = 5.0J