The energy stored in a spring of stiffness constant k = 2000(Nm^{-1}) when extended 4cm is
-
A.
0.16J -
B.
1.60J -
C.
16.00J -
D.
160.00J
Explanation
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)