## A bungee cord is 30m long and when stretched a distance x itexerts a restoring force of magnitude kx. Your father-in-law (mass95.0kg) stands

Question

A bungee cord is 30m long and when stretched a distance x itexerts a restoring force of magnitude kx. Your father-in-law (mass95.0kg) stands on a platform 45.0m above the ground , and one endof the cord is tied securley to his ankle and the other end to theplatform. You have promised him that when he steps off he will falla maximum of 41m before the cord stops him. You had several bungeecords to select from and tested them by stretching them out, tyingone end to a tree and pulling the other end with a force of 380.0N.When you did this, what distance will the bungee cord you shouldselect have stretched?

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1 year 2021-09-03T04:06:37+00:00 1 Answers 0 views 0

## Answers ( )

The distance the bungee cord that would be stretched 0.602 m, should be selected when pulled by a force of 380 N.

Explanation:

As from the given data

the length of the rope is given as l=30 m

the stretched length is given as l’=41m

the stretched length required is give as  y=l’-l=41-30=11m

the mass is m=95 kg

the  force is  F=380 N

the gravitational acceleration is g=9.8 m/s2

The equation of  k is given by equating the energy at the equilibrium point which is given as

$$U_{potential}=U_{elastic}\\mgh=\dfrac{1}{2} k y^2\\k=\dfrac{2mgh}{y^2}$$

Here

m=95 kg, g=9.8 m/s2, h=41 m, y=11 m so

$$k=\dfrac{2mgh}{y^2}\\k=\dfrac{2\times 95\times 9.8\times 41}{11^2}\\k=630.92 N/m$$

Now the force is

$$F=kx\\$$ or

$$x=\dfrac{F}{k}\\$$

So here F=380 N, k=630.92 N/m

$$x=\dfrac{F}{k}\\x=\dfrac{380}{630.92}\\x=0.602 m$$

So the distance is 0.602 m