How Far Will Dave Fall Before the Bungee Cord Stops Him?

AI Thread Summary
Dave jumps off a bridge with a bungee cord, falling 15 meters before the cord stretches. His mass is 75 kg, and the bungee cord follows Hook's law with an elastic constant of 50 N/m. To find how far he falls before stopping, the gravitational potential energy (GPE) and elastic potential energy (EPE) are equated, leading to the conclusion that he falls 55 meters in total. While some suggest using force calculations, the energy approach is more straightforward since both force and extension are unknowns. The discussion emphasizes the importance of conservation of energy in solving the problem.
Delzac
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hi i have a Qns.

Dave jumps off a bridge with a bungee cord( a stretchable cord) tied around his ankle. He falls for 15m before the bungee cord begins to stretch. Dave's mass is 75kg and we assume the cord obeys Hook's law with elastic constant, k= 50N/m. if we neglect air resistance and the mass of the cord, calculate how far below the bridge Dave will fall before coming to a stop.

i understand that i have to use the formula F=Ke and i need to find F.
how should i go about finding it? which formula to use?
 
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Consider conservation of energy.
 
Got it!

Let X be the height Dave fall before comin to a stop
Let e be the extension of bungee cord.

GPE=mgh
=(75)(9.82)(X)

Elastic Potential Energy= 1/2Ke^2
= 1/2(50)(X-15)^2

Equate it together and X = 55m.

However can't u sovle by finding Force??
 
Delzac said:
However can't u sovle by finding Force??
There's no direct way to use F=ke to solve for e (if that's what you mean), since both F and e are unknowns.
 
Delzac said:
Got it!

Let X be the height Dave fall before comin to a stop
Let e be the extension of bungee cord.

GPE=mgh
=(75)(9.82)(X)

Elastic Potential Energy= 1/2Ke^2
= 1/2(50)(X-15)^2

Equate it together and X = 55m.

However can't u sovle by finding Force??
You are indirectly using force when you use energy. The incremental potential energy stored in the cord is dU = Fde, so total potential energy is:

U = \int_{0}^{e_f}Fde = \int_{0}^{e_f}Kede = \frac{1}{2}Ke_f^2

[Note: Ke is spring constant x extension, not Kinetic energy]

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