An 8m long copper wire and a 4m long steel wire, each of cross-section 0.5 cm2 are fastened end to end and stretched by a 500 N force. The elastic potential energy of the system is (YCopper=1×1011Nm−2;Ysteel=2×1011Nm−2)
The potential energy of an elongated rod is given by U=F2L2AY. In this problem the total
potential energy is the sum of the individual potential energies. Therefore,
U=F22A(L1Y1+L2Y2)
On substitution,
U=0.25J=14J