Hi, A rather simple question that I just can't seem to solve: The greatest instantaneous acceleration a person can survive is 25g, where g is the acceleration of free fall. A climber's rope should be selected such taht, if the climber falls when the rope is attached to a fixed point on a vertical rock, the fall will be survived. A climber of mass m is attached to a rope which is attached firmly to a rock face at B as shown. When at a point A, a distance L above B, the climber falls. http://img144.imageshack.us/img144/2483/rockclimbjh9.jpg [Broken] (the red bit is the rope) (a) Assuming that the rope obeys Hooke's law upto the breaking, use the principle of conservation of energy and the condition for greatest instantaneous acceleration to show that the part AB of the rope (of unstretched length L) must be able to stretch by more than L/6 whithout breaking for the climber to survive My working: After the climber falls under freefall for a maximum distance of 2L (from A to B, and then the length L of the rope which makes it a total of 2L), then he has maximum kinetic energy which is K.E = [tex] (1/2)mv^2 [/tex] but [tex] v^2 = u^2 + 2as = 4gL [/tex] Then K.E. = elastic potential energy + gpe ... (i) EPE = [tex] (1/2)kx^2 = (1/2)Fx [/tex] and GPE = mgx , where x is the extension during the rope stretching. So from (i): [tex] (1/2)mv^2 = mgx + (1/2)Fx [/tex] and since F = ma, with the maximum instantaneous acceleration of 25g, then F = 25mg [tex] 2mgL = mgx + (25/2)mgx [/tex] [tex] 2L = (27/2)x [/tex] Therefore [tex] x = (4/27) L [/tex] ? but the required answer is x = L/6 ?