An elastic cable is to be designed for bungee jumping from a tower 130 ft high. The specifications call for the cable to be 85 ft long when unstretched, and to stretch to a total length of 100 ft when a 750-lb weight is attached to it and dropped from the tower. Determine:
a. The required spring constant k of the cable.
b. How close to the ground a 185-lb man will come if he uses this cable to jump from the tower?

a) The spring constant is 50 lb/ft

b) The man is 26.3 ft close to the ground.

Explanation:

Height of tower is 130 ft

Specification calls for a cable of length 85 ft

the maximum this length stretches is 100 ft when subjected to a load of 750 lb

The extension of the cable is calculated from the formula from Hooke's law

F = kx

where F is the load or force on the cable

k is the spring constant of the cable

x is the extension on the cable

a) The extension on the cable is

x = 100 ft - 85 ft = 15 ft

substituting into the formula above, we'll have

750 = k*15

k = 750/15 = 50 lb/ft

b) for a 185 lb man, jumping down will give an extension gotten as

F = kx

185 = 50*x

x = 185/50 = 3.7 ft

The total length of the cable will be extended to 100 ft + 3.7 ft = 103.7 ft

closeness to the ground = 130 ft - 103.7 ft = 26.3 ft

answer: 500 wph.

explanation:

answer:

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explanation: