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4 years ago

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A 70.0 kg stuntman is attached to a bungee cord with an unstretched length of 15.0 m. He jumps off a bridge spanning a river fro

m a height of 50.0 m. When he finally stops, the cord has a stretched length of 44.0 m. Treat the stuntman as a point mass, and disregard the weight of the bungee cord. Assuming the spring constant of the bungee cord is 71.8 N/m, what is the total potential energy relative to the water when the man stops falling?

1 answer:

xz_007 [3.2K]4 years ago

7 0

Answer:

The total potential energy is 34.307 kJ

Solution:

As per the question:

Mass of the stuntman, m = 70.0 kg

Length of the unstretched cord, l = 15.0 m

Height of the cord, H = 50.0 m

The stretched length of the cord, $\Delta l = 44.0\ m$

Spring constant of the cord, k = 71.8 N/m

Now,

The total potential energy is given as the sum of gravitational potential energy and elastic potential energy:

The height attained, h = H - $\Delta l$ = 50.0 - 44.0 = 6.0 m

PE_{G} = mgh = $70\times 9.8\times 6.0 = 4116\ J = 4.116\ kJ$

Now, the elastic potential energy:

x = l - $\Delta l$ = 44.0 - 15.0 = 29.0 m

$PE_{E} = \frac{1}{2}kx^{2} = \frac{1}{2}\times 71.8\times (29)^{2} = 30.191\ kJ$

The Total Potential Energy:

$PE_{G} + PE_{E} = 4.116 + 30.191 = 34.307\ kJ$

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