An 8,m long copper wire and 4,m long steel wi | vedclass

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(A) The elastic potential energy stored in a stretched wire is given by $U = \frac{1}{2} F \Delta l$.Since the wires are connected in series,the same

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$1/4\,J$

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(A) The elastic potential energy stored in a stretched wire is given by $U = \frac{1}{2} F \Delta l$.Since the wires are connected in series,the same force $F = 500\,N$ acts on both.The cross-sectional area $A = 0.5\,cm^2 = 0.5 	imes 10^{-4}\,m^2$.Extension in copper wire: $\Delta l_1 = \frac{F L_1}{Y_{cu} A} = \frac{500 	imes 8}{1 	imes 10^{11} 	imes 0.5 	imes 10^{-4}} = \frac{4000}{0.5 	imes 10^7} = 8 	imes 10^{-4}\,m = 0.8\,mm$.Extension in steel wire: $\Delta l_2 = \frac{F L_2}{Y_{steel} A} = \frac{500 	imes 4}{2 	imes 10^{11} 	imes 0.5 	imes 10^{-4}} = \frac{2000}{1 	imes 10^7} = 2 	imes 10^{-4}\,m = 0.2\,mm$.Total extension $\Delta l = \Delta l_1 + \Delta l_2 = 0.8\,mm + 0.2\,mm = 1.0\,mm = 10^{-3}\,m$.Total elastic potential energy $U = \frac{1}{2} F \Delta l = \frac{1}{2} 	imes 500 	imes 10^{-3} = 0.25\,J = 1/4\,J$.

An $8\,m$ long copper wire and $4\,m$ long steel wire,each of cross-section $0.5\,cm^2$,are fastened end to end and stretched by a $500\,N$ force. The elastic potential energy of the system is (Young's modulus: $Y_{cu} = 1 \times 10^{11}\,N/m^2$,$Y_{steel} = 2 \times 10^{11}\,N/m^2$):

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