The energy stored in a spring of stiffness constant k = 2000\(Nm^{-1}\) when extended 4cm is
0.16J
1.60J
16.00J
160.00J
Correct answer is B
The spring constant = the measure of stiffness = 2000\(Nm^{-1}\)
Energy stored in a spring = Work done to stretch it through a distance = change in potential energy
= \(\frac{kx^{2}}{2}\) = \(\frac{2000 \times (0.04)^{2}}{2}\) (converting 4cm to metres)
= \(\frac{3.2}{2} = 1.60J\)