An 85.0 kg bungee jumper jumps from a bridge 120.0 m above the canyon floor. The

Question

An 85.0 kg bungee jumper jumps from a bridge 120.0 m above the canyon floor. The Bungee cord, which obeys Hooke's Law, has a relaxed length of 60.0 m and a spring constant of 62.3 N/m. Ignore air resistance and the jumper's height for the following questions, and use g = 10 m/s2.

a. What is the jumper's speed just before the bungee cord begins to stretch?

b. How far above the canyon floor will the jumper come to a stop?

c. At the bottom of his jump, what is the net force (magnitude and direction) acting on the bungee jumper?

d. After "rebounding," bouncing a few times, and finally coming to rest, how far is the bungee cord stretched beyond its relaxed length?

e. From this post-jump equilibrium point, a host mechanism is used to raise the jumper back to the bridge in 30 seconds. What is the minimum horsepower rating of the motor? (1 hp = 746 W)

f. What is the maximum "safe" mass for a jumper using this cord and jumping from this bridge (in other words, what mass must the jumper be to just touch the canyon floor with a speed of 0 m/s when he does)?

Explanation / Answer

a) Apply energy conservation

0.5*m*v^2 = m*g*L

v = sqrt(2*g*L)

= sqrt(2*10*60)

= 34.64 m/s

b)Again Apply energy conservation

let x is the extension of the cord
0.5*k*x^2 = m*g*(L+x)

0.5*62.3*x^2 = 85*10*(60+x)

31.15*x^2 = 51000 + 850*x

x^2 -27.3*x - 1632 = 0

solving bove equation

x = 56.3 m

so final height of the floor, h = 120 - (60+56.3)

= 3.7 m

c) Fnet = k*x - m*g

= 62.3*56.3 - 85*10

= 2658 N (upward)

d) Fnet = 0

k*x - m*g = 0

x = m*g/k

= 85*10/62.3

= 13.64 m

e)

power = Workdone/time

= m*g*(L+x)/t

= 85*10*(60+13.64)/30

= 2086.5 Watts

= 2.797 hp

f)

0.5*62.3*60^2 = m*10*(60+60)

m = 93.45 kg

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