The stress experimented by the bungee cord due to Max's weight is approximately 299.680 kilopascals.
Let suppose that the stress ([tex]\sigma[/tex]), in pascals, experimented by the bungee cord is entirely elastic and uniform. By Newton's Laws we know that the tension ([tex]T[/tex]), in newtons, experimented by the cord equals Max's weight. The stress on the bungee cord is axial and is described by following formula:
[tex]\sigma = \frac{m\cdot g}{\frac{\pi\cdot D^{2}}{4} }[/tex] (1)
Where:
If we know that [tex]m = 15\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]D = 0.025\,m[/tex], then the stress experimented by the bungee cord is:
[tex]\sigma = \frac{(15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{\frac{\pi}{4}\cdot (0.025\,m)^{2} }[/tex]
[tex]\sigma \approx 299679.845\,Pa\, (299.680\,kPa)[/tex]
The stress experimented by the bungee cord due to Max's weight is approximately 299.680 kilopascals. [tex]\blacksquare[/tex]
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