A cat dozes on a stationary merry-go-round, at a radius of 4.6 m from the center of the ride. then the operator turns on the rid
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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
Here
m=95 kg, g=9.8 m/s2, h=41 m, y=11 m so
Now the force is
or
So here F=380 N, k=630.92 N/m
So the distance is 0.602 m