Answer:
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
[tex]U_{potential}=U_{elastic}\\mgh=\dfrac{1}{2} k y^2\\k=\dfrac{2mgh}{y^2}[/tex]
Here
m=95 kg, g=9.8 m/s2, h=41 m, y=11 m so
[tex]k=\dfrac{2mgh}{y^2}\\k=\dfrac{2\times 95\times 9.8\times 41}{11^2}\\k=630.92 N/m[/tex]
Now the force is
[tex]F=kx\\[/tex] or
[tex]x=\dfrac{F}{k}\\[/tex]
So here F=380 N, k=630.92 N/m
[tex]x=\dfrac{F}{k}\\x=\dfrac{380}{630.92}\\x=0.602 m[/tex]
So the distance is 0.602 m
