A daredevil plans to bungee jump from a balloon 76.0 m above the ground. He will

Question

A daredevil plans to bungee jump from a balloon 76.0 m above the ground. He will use a uniform elastic cord, tied to a harness around his body, to stop his fall at a point 11.0 m above the ground. Model his body as a particle and the cord as having negligible mass and obeying Hooke's law. In a preliminary test he finds that when hanging at rest from a 5.00-m length of the cord, his body weight stretches it by 1.45 m. He will drop from rest at the point where the top end of a longer section of the cord is attached to the stationary balloon.

(a) What length of cord should he use? m

(b) What maximum acceleration will he experience? m/s2

Explanation / Answer

a)

here

k = F / x

k = M * 9.8 / 1.45 = 6.75M N/kg m

initailly the totalenergy is = PE = m * g * h = M * 9.8 * 76 = 744.8 M m^2/s^2

at maximum stretch the energy is = PE + U = m * g * h + 0.5 * k * x^2

744.8 * M = M * 9.8 * 11 + 0.5 * 6.75 * M * x^2

x = 13.738 m

L = (76 - 11) - 13.738 = 51.26 m

b)

F = k * x = 6.75M * 13.738 = 92.73M N/kg

then

F = m*a

a = F / m

a = 92.73M / M = 92.73 m/s^2

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