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PROBLEM 2.4\n\nKNOWN: A soap film on a wire frame is stretched.\nFIND: Determine the work done.\nSCHEMATIC & GIVEN DATA:\n\ndrag model: (1) The film is a closed system. (2) The moving boundary is the only work mode. (3) The surface tension is constant, acting on both sides of the film.\nANALYSIS: (a) The work is determined using Eq. 2.19\nw = \\int_{x_1}^{x_2} c_A = - \\int_{x_1}^{x_2} 2 T dX\nFor constant surface tension T\nW = 2 T Δx\n\n(b) If r = 5cm, Δx = 0.5cm, T = 2.5x10^-5 N/cm\n\nW_f = (2.5x10^-5 N/cm)(2)(5cm)(0.5cm)\\left| \\frac{1 mm}{0.1 cm} \\right|\\left| \\frac{1 J}{1 N-m} \\right|= -125 x10^-5 J\n\nThe negative sign denotes work done on the film. Note the small magnitude of the work required to stretch the film.\n\nPROBLEM 2.42\n\nKNOWN: Data is provided for a spring stretched by a force applied at its end.\nFIND: Obtain an expression for the work done in stretching the spring and evaluate the work using given data.\nSCHEMATIC & GIVEN DATA:\n\nENGR MODEL:\n1. The spring is the closed system.\n2. The moving boundary is the only work made.\n3. Hooke's law applies.\nANALYSIS: All the work done in stretching the spring is given by\nW = \\int F dx\nLetting x = x_f - x_0, this becomes\nW = - \\int_{x_0}^{x_f} k (x_f - x) dx = - \\frac{1}{2} k \\left( x_f^2 - x_0^2 \\right)\n\n(b) When (x_i - x_0) = 3cm and (r_0 - r_l) = 1cm,\nW_f = \\left( -10^{-4} N \\right) \\left[ (x_f - x_0)^{-2} (10^{-4})^2 \\right] \\left| \\frac{1 mm}{0.1 cm} \\right|\\left| \\frac{1 J}{1 N-m} \\right| = - 20 J