# physics oscillation question! ?

A 85 kg student jumps off a bridge with a 9-m-long bungee cord tied to his feet. The massless bungee cord has a spring constant of 390 N/m. You can assume that the bungee cord exerts no force until it begins to stretch.

Part A. How far below the bridge is the student's lowest point?

Part B. How far below the bridge is the student's resting position after the oscillations have been fully damped?

PLEASE HELP!!

### 3 Answers

- NCSLv 71 month ago
A. Let's say the spring stretches a maximum of x.

GPE has become SPE:

mg(9 + x) = ½kx²

85*9.8*(9+x) = ½*390*x²

This quadratic has a negative root we can ignore

and a positive root at x = 8.7 m

Max distance below bridge is 9 + 8.7 = 17.7 m

which one might argue should be rounded to 20 m owing to the data.

B. At equilibrium, spring force = weight

kx = mg

x = mg / k = 1.5 m

so equilibrium distance below bridge is 9 + 1.5 = 10.5 m

(≈ 10 m)

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- VamanLv 71 month ago
Up ti 9 meters, it is free fall. Kinetic energy acquired is m g h. m is the mass=85 kg, g=9.8, h=9, Energy acquired=9497 J. This is balanced by the rope. It extends by x meters. We have 9497 = 1/2 *390*x^2, x^2=4=38.45. x=6.2 meters. This is the extension. The lowest point=9+6.2=15.2 m. Ans B is also the same.

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- Andrew SmithLv 71 month ago
b) 9 + kx = mg ( the forces are in equilibrium.. solve for x)

a) 9 + 1/2 k x^2 = m g (9+x) where x is the amount of stretch once the cord is taut.

-> 195 x^2 - mgx +9(1-mg) = 0 Substitute and solve the quadratic.

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You're right. Should be 2.1 m, plus the 9 m makes ≈ 11 m