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A climber of mass m is scaling the vertical wall of a mountain. His safety rope is made of nylon that, when stretched, behaves like a sping with a spring constant k. He stops to rest, but then accidentally slips and falls freely for a distance h before the rope runs out of slack. The rope then stretches an amount x as it breaks his fall and momentarily brings him to rest. Derive an expression for the climbers mass, m, in terms of k, h, x, and g (the acceleration due to gravity).
When i think about this i automatically think to use the equation:
1/2(k)(y final^2) = (m)(g)(h initial)...
but I do not think that you can use two distances, it wouldn't make sense. Or is the y = to the height?
When i think about this i automatically think to use the equation:
1/2(k)(y final^2) = (m)(g)(h initial)...
but I do not think that you can use two distances, it wouldn't make sense. Or is the y = to the height?