Question

Design a "bungee jump" apparatus for adults. A bungee jumper falls from a high platform with two elastic cords tied to the ankes. The jumper falls freely for a while, with the cords slack. Then the jumper falls an additional distance with the cords increasingly tense. You have cords that are 10m long, and these cords stretch an additional 24m for a jumper whose mass is 80kg, the heaviest adult you will allow to use your bungee jump (heavier customers would hit the ground). You can neglect air resistance. (a) Make a series of five simple diagrams, like a comic strip, showing the platform, the jumper, and the two chords at various times in the fall and the rebound. On each diagram, draw and label vectors representing the forces acting on the jumper, and the jumper's velocity. Make the relative lengths of the vectors reflect their relative magnitudes. (b) At what instant is there the greatest tension in the chords? How do you know? (c) What is the jumper's speed at this instant? (d) Is the jumper's momentum changing at this instant or not? (That is, is dp/dt nonzero or zero) Explain briefly. (e) Focus on this instant, and use the principles of this chapter to determine the spring stiffnes, k_s for each cord. Explain your analysis. (f) What is the maximum tension that each cord must support without breaking? (g) What is the maximum acceleration (in g's) that the jumper experiences? What is the direction of this maximum acceleration? (h) State clearly what approximations and estimates you have made in your design.

Answer 1

a )

the cord is slack until 10 m extended

here at this the constant gravity acts on the jumper , after this it will be at equilibrium , the elastic force the variable

acts on jumper it to bring back him into equilibrium, after above the equlibrium he undergoes harmonic motion , not SHM

b )

elastic force is higher when he is in downward form the equilibrium below 24 m

under Hooke's law and F is proportional to x

c )

at that instant speed is zero

the upward and downward the velocity changes

d )

momentum is changing so dp/dt is not equal to zero

so it is above

e )

using equation

m g ( h - x ) = k x²

k = 80 X 9.8 X ( 10 + 24 ) / 24²

k = 46.277 N/m

F = k x

F = 46.277 X 24

F = 1110.66 N

h )

no air resistance and the elastic obey's Hooke's law