The energy stored in a spring of stiffness constant k = 2000\(Nm^{-1}\) when extended 4cm is

The energy stored in a spring of stiffness constant k = 2000\(Nm^{-1}\) when extended 4cm is

Answer Details

The energy stored in a spring can be calculated using the formula: Energy (E) = 0.5 * k * x^2 where k is the stiffness constant of the spring and x is the amount it is extended or compressed from its rest position. In this case, k = 2000 Nm^-1 and x = 0.04 m (4 cm converted to meters). Substituting the values into the formula: E = 0.5 * 2000 * 0.04^2 E = 0.5 * 2000 * 0.0016 E = 1.60 J So, the energy stored in the spring when it is extended 4 cm is 1.60 J.