A 57.0 kg skier is being pulled up a 13.0° slope by a steelcable. The cable has a cross-sectional area of 7.18 10-5 m2. The cable applies a force tothe skier, and, in so doing, the cable stretches by 2.5110-4 m. A frictional force of 68.0 N acts on theskis and is directed opposite the skier's motion. If the skier'sacceleration up the slope has a magnitude of 1.04 m/s2,what is the original unstretched length of the cable?
F = Y (ΔL / L) A
F = 2.0x10^11 N/m^2 (2.51 x 10^-4 m / L) 7.18 x 10^-5m^2
L = ?