# Challenging Qns On Work Energy Power

#### Delzac

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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#### Doc Al

Mentor
Consider conservation of energy.

#### Delzac

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??

#### Doc Al

Mentor
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.

#### Andrew Mason

Homework Helper
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]

AM

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