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A bungee jumper of mass m jumps off a bridge. Assume that the bungee cord behaves like am ideal spring of spring constant k. When the jumper is at the lowest point in the motion that follows, the extension of the bungee cord is x.
After reaching the lowest point for the first time, the jumper rebounds straight upwards.
What is the magnitude of her velocity at the instant that the cord is at its natural length?

a) √(k/m) x
b)√[(kx²/m) - 2gx]
c) √(2gx)
d) √(-2gx)
e) None of the above

Respuesta :

rafaleo84

Answer:

b) √[(kx²/m) - 2gx]

Explanation:

The energy at the lowest point is equal to:

[tex]E_{elas}=\frac{1}{2} *k*x^{2}[/tex]

where:

Eelas = elastic energy [J]

k = spring constant [N/m]

x = extension of the spring [m]

We consider the lowest point, as the point where the potential energy is zero. At the moment when the person goes back through the point of the normal length of the elastic cord, it is this point that the person will have potential energy and kinetic energy.

[tex]E_{elas}=E_{pot}+E_{kine}\[/tex]

[tex]\frac{1}{2}*k*x^{2}=m*g*x +\frac{1}{2} *m*v^{2} \\v^{2} = \frac{k*x^{2} }{m}-2*g*x\\ v=\sqrt{\frac{k*x^{2} }{m}-2*g*x}[/tex]

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