Question

A 58.0 kg bungee jumper is standing on a tall platform (ho = 46.0 m) as shown in the diagram. The bungee cord has an unstrained length of Lo = 9.00 m, and when stretched, behaves like an ideal spring with a spring constant of k = 61.0 N/m. The jumper falls from rest, and the only forces acting on him are gravity 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 = 37.0 m and (b) hB = 15.0 m? Ignore the mass of the bungee cord.

Answer 1

Height of bridge = 46m Rope Length = 9m Mass = 68 kg To find elongation force in (F=kx) F=m*g F=(75kg)x(9.81m/s^2) F=735.75N F=k*x

mass = 75kg height, before = 200m, after = 140m Energy before = Energy After Potential + Kinetic = Potential + Kinetic mgh + 0.5mv^2 = mgh + 0.5mv^2 (75)(9.81)(200) +(0.5)(75)(0^2) = (75)(9.81)(140) + (0.5)(75)(v^2) 147150J + 0 = 103005J + 37.5v^2 44145J = 37.5v^2 v^2 = 1177.2 v = 34.31m/s This is the velocity at 140m (a 60m dive)... I think. Can anyone see a flaw in my calculations? velocity = displacement / time time = displacement / velocity time = 60m / 34.31m/s time = 1.75 s The jumper is 40m above ground at 1.75 seconds. Now for when the rope has elastic potential energy at maximum displacement: Energy Before = Energy After 147150J = Potential + Kinetic + Elastic 147150J = mgh + 0.5mv^2 + 0.5kx^2 147150J = (75)(9.81)(h) + 0.5(75)(0^2) + 0.5(k)(x^2) We need the height, the constant in the rope, and the maximum displacement,x, Using F=kx and F=mg to determine k. mg=kx (75)(9.81)=kx The maximum elasticity of the rope is 50%, so a rope that is 60m will stretch a maximum of 30m, before it loses it's elasticity (Hooke's Law). So, x = 30m. 735.75=-k(30) k=-24.525N/m 147150J = (75)(9.81)(h) + 0.5(75)(0^2) + 0.5(-k)(30^2) 147150J = (735.75)(h) + 0 + -11036.25 735.75h =158186.25 h = 215m