A 70.5-kg bungee jumper is standing on a tall platform (h0 = 49.6 m), as indicat

Question

A 70.5-kg bungee jumper is standing on a tall platform (h0 = 49.6 m), as indicated in the figure. The bungee cord has a natural length of L0 = 10.0 m and, when stretched, behaves like an ideal spring with a spring constant of k = 51.5 N/m. The jumper falls from rest, and it is assumed that the only forces acting on him are his weight and, for the latter part of the descent, the elastic force of the bungee cord. What is his speed when he is at the following heights above the water: (a) hA = 39.6 m, and (b) hB = 14.9 m?

Explanation / Answer

initial potnetial energy = m*g*ho

final potential energy = m*g*hA

final kinetic energy = Kf = 0.5*m*v^2

from energy conservation

m*g*ho = m*g*hA + 0.5*m*v^2

(70.5*9.8*49.6) = (70.5*9.8*39.6)+(0.5*70.5*v^2)

v = 14 m/s

(b)

final potnetial energy = m*g*hB

elestic potnetial energy = 0.5*k*dy^2

kinetic energy = 0.5*m*v^2

dy = (49.6-10-14.9) = 24.7 m

(70.5*9.8*49.6) = (70.5*9.8*14.9)+(0.5*70.5*v^2) + (0.5*51.5*24.7^2)

v = 15.31 m/s

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