How to find the Max. distance of the bungee jumper?

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To find the maximum distance of a bungee jumper's fall, the conservation of mechanical energy principle must be applied. The total mechanical energy before the jump equals the total mechanical energy after the jump, represented by the equation K_1 + U_grav,1 = K_2 + U_grav,2 + U_el,2. The gravitational potential energy at the start is mgL, while the elastic potential energy must be calculated using the spring constant related to the elastic modulus of the bungee cord. The discussion emphasizes the need to derive an equation connecting the spring constant to the elastic modulus and solve for the maximum distance fallen. Understanding these energy transformations is crucial for solving the problem effectively.
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Please help me to solve this question because I don't have a clue to do that!

Q: In a bungee jump a volunteer of mass 70 kg drops from a bridge, tethered to his jump point by an elastic cable of unstretched length L = 20m and elastic modulus 3000 N. Ifnoring energy losses, and assuming he hits nothing below, find the jumper's maximum distance of fall.

Thanks!
 
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HINT:Use the law of conservation of mechanical energy.Pay attention with the 3 various types of energy the system has and with the "zero" for gravitational potential energy.

Daniel.

P.S.The problem is posted twice...
 
Esta said:
Please help me to solve this question because I don't have a clue to do that!

Q: In a bungee jump a volunteer of mass 70 kg drops from a bridge, tethered to his jump point by an elastic cable of unstretched length L = 20m and elastic modulus 3000 N. Ifnoring energy losses, and assuming he hits nothing below, find the jumper's maximum distance of fall.

Thanks!

The total mechanical energy before the jump is equal to the total mechanical energy after the jump.

The relevant equation is K_1 + U_{grav,1} = K_2 + U_{grav,2} + U_{el,2}.

\frac{1}{2}mv_1^{2} + mgy_1 = \frac{1}{2}mv_2^{2} + mgy_2 + \frac{1}{2}ky_2^{2}

Choosing the relaxed hanging length of the rope as the origin,

0 + mgL = mgy_2 + \frac{1}{2}ky_2^{2}.

Exercise for the reader: find an equation relating the spring constant to the elastic modulus of the bungee cord and solve the above equation for y_2. Hint: the distance of fall is not y_2.
 
Potential energy for spring constant F=- kx, k=λ*A*x/L λ; however, now (A)cross-section area of the cord is provided; for string constant F = -λ (x/L). But will they equal?! kx = λ (x/L)?!

Sorry, still not very understand! :frown:
 
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