A 100.0 kg bungee jumper steps off a bridge with a light bungee cord tied to her
ID: 1410779 • Letter: A
Question
A 100.0 kg bungee jumper steps off a bridge with a light bungee cord tied to her and to the bridge. The unstretched length of the cord is 10.0 m. She reaches the bottom of her motion 35.0 m below the bridge before bouncing back. Her motion can be separated into an 10.0 m free-fall and a 25.0 m section of simple harmonic oscillation.
(a) For what time interval is she in free-fall? (units: in seconds)
(b) Use the principle of conservation of energy to find the spring constant of the bungee cord. (units: N/m)
(c) What is the location of the equilibrium point where the spring force balances the gravitational force acting on the jumper? Note that this point is taken as the origin in our mathematical description of simple harmonic oscillation. (units: meters, below the bridge)
(d) What is the angular frequency of the oscillation? (units: rad/s)
(e) What time interval is required for the cord to stretch by 25.0 m? (units: s)
(f) What is the total time interval for the entire 35.0 m drop? (units: s)
Explanation / Answer
E) This problem is best broken up into three parts
1. Free fall
2. time between free fall and oscillation
3. Oscillation
(we will use these numbers to say where we are and velocity at the end of these 3 periods)
1) V1^2=Vo^2+2a1d -> V1= Sqrt(2*a1*d) = sqrt(2(9.8)(11)) -> V1 = 14.683 m/s
V1 = Vo +(a1)(t1) -> 14.683 / 9.8 = t1 = 1.498 s
2) We know the bungee cable engages at 11 meters below the bridge we know that at this time our Velocity is 14.683 m/s but we need to know what our velocity is at the new equilibrium of the system so we use kinematics.
x2 = (d2-d1) = the amount that the bungee cord has stretched from 11 feet below the bridge to the new equilibrium of the system
Uo = mg(delta y) or -mg(d2-do) = 100(-9.8)(19.68 - 35) = 9596.16 J
U2 = (1/2)k(d2-d1)^2 -> (1/2)(67.81)(8.68)^2 = 2554.48 J
K2 = (1/2)MV2^2
energy is conserved to this point so 9596.46 = 2554.48 + (1/2)MV2^2
V2^2 = 2(9596.46 - 2554.48) / M -> V2^2 = 234.723 -> V2 = 15.32 m/s
V2^2 = V1^2 + 2(a2)(d2-d1) -> 15.32^2 - 14.683^2 = 2(a2)(8.68) -> a2 = 1.112
V2 = V1 + a(t2) -> (15.32 - 14.683) / 1.112 = t2 = .578 s
3) We start from equilibrium and go 1/4 rotation to reach the bottom so pi/2
T = 2pi / w so 1/4T = pi / 2w so t3 = pi / 2(1.06) = 1.482 s
To answer E we need to just add up t2 and t3 and we get 2.06 s
F) to get F all we do is add up t1 t2 and t3 to get the overall time which is 3.558 s
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