1. A rival football player is listed on the roster at 240 lbs, but you’re pret

Question

1. A rival football player is listed on the roster at 240 lbs, but you’re pretty sure it’s

an exaggeration. At a pool party, you observe him bobbing up and down on the end

of a diving board. It occurs to you that the end of the diving board exhibits periodic

motion, which you know all about, and you devise a plan to determine his weight.

You measure that he bobs up and down 5 times in 6.0 seconds, and determine that

when you hang on to the end of the board it bends down by 15 cm. Knowing that

you weigh 600 N (135 lbs), what’s your friend’s true weight? (Hint: Model the end of

the diving board as a spring.)

2. In bungee jumping, an operator attaches an elastic (spring-like)

cord to a jumper. The jumper then jumps off a bridge. She falls

freely until she reaches the unstretched length of the cord. Then

the cord begins to stretch and slow her to a stop. It then yanks her

back up and she oscillates up and down for a while. When she is

nearly stopped, the operator pulls on the elastic cord and hauls her

back up to the bridge.

a. How fast is she falling when she reaches the length of the

unstretched cord? Let’s say the bridge is 75 m high, and the length

of the unstretched cord is 8 m.

b. If the rope stretches 4 m to slow her to a stop, what’s the spring

constant of the elastic cord? Take her mass to be 50 kg.

c. About how long will it take her to oscillate up and down once?

Explanation / Answer

This is a slightly modified answer I gave another person a couple of hours back, same question but different wording.
1) Spring constant (K) = 600/0.15m = 4,000N/m.
(6/5) = frequency of 1.2Hz.
Period (T) = (1/F) = 0.833 sec.
T = 2pi sqrt.(M/K).
Sqrt.(M/K) = T /2Pi.
(M/K) = (T/2pi)^2.
(T/2pi)^2 x K = M.
M = 70.28742634kg., weight = 688.82N. (9.8 used as g).

2)
(a) In 8m. the jumper reaches a speed of sqrt.(2gh) = 12.522m/sec.
She stops in a distance of 4m. from 12.522m/se., so acceleration = (v^2/2d) = -19.6m/sec^2.
Force = (ma) = 50 x 19.6, = 980N.
b) The cord has stretched 4m. with a force of 980N., so constant K = (980/4) = 245N/m.
c) Frequency = sqrt.((K/M)/2pi) = 0.883 Hz.
Time = (1/f) = 1.134 secs.

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