Physics for Scientists and Engineers
Physics for Scientists and Engineers
10th Edition
ISBN: 9781337553278
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Textbook Question
Chapter 15, Problem 14P

Review. A 65.0-kg bungee jumper steps off a bridge with a light bungee cord tied to her body and to the bridge. The outstretched length of the cord is 11.0 m. The jumper reaches the bottom of her motion 36.0 m below the bridge before bouncing back. We wish to find the time interval between her leaving the bridge and her arriving at the bottom of her motion. Her overall motion can be separated into an 11.0-m tree fall and a 25.0-m section of simple harmonic oscillation. (a) For the free-fall part, what is the appropriate analysis model to describe her motion? (b) For what time interval is she in free fall? (c) For the simple harmonic oscillation part of the plunge, is the system of the bungee jumper, the spring, and the Earth isolated or nonisolated? (d) From your response in part (c) find the spring constant of the bungee cord. (c) What is the location of the equilibrium point where the spring force balances the gravitational force exerted on the jumper? (f) What is the angular frequency of the oscillation? (g) What time interval is required for the cord to stretch by 25.0 m? (h) What is the total time interval for the entire 36.0-m drop?

(a)

Expert Solution
Check Mark
To determine

The appropriate analysis model of jumper’s motion for the free fall part.

Answer to Problem 14P

The jumper’s motion has the constant acceleration.

Explanation of Solution

A free falling object is an object that is falling under the sole influence of gravity.

Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. The free fall phase follows the parabolic behavior. Since the only gravity acted on the free fall, the acceleration is constant.

Conclusion:

Therefore, the jumper’s motion has the constant acceleration.

(b)

Expert Solution
Check Mark
To determine

The time required for free fall.

Answer to Problem 14P

The time required for free fall is 1.49s.

Explanation of Solution

The mass of bungee jumper is 65.0kg, the stretched length of the cord and free fall height is 11.0m, the distance where the jumper reaches before bouncing back is 36.0m and the height of simple harmonic oscillation is 25.0m.

The equation for the kinematic is,

    yf=yi+vyt+12ayt2

Here, yf is the free fall height, t is the time, yi is the initial height, vy is the velocity and ay is the acceleration.

Substitute 11.0m for yf, 0 for yi, 0 for vy and 9.8m/s2 for ay in above equation to find t.

  11.0m=0+(0)t+12(9.8m/s2)t2=12(9.8m/s2)t2t=(2×11.0m)9.8m/s2=1.49s

Conclusion:

Therefore, the time required for free fall is 1.49s.

(c)

Expert Solution
Check Mark
To determine

Weather the system of the bungee jumper, the spring and the earth is isolated or non-isolated for simple harmonic oscillation.

Explanation of Solution

When a system is isolated, it means that it is separated from its environment in such a way that no energy flows on or out of the system. The non-isolated system interacts with its environment and exchanges the energy.

The energy of the system of the bungee jumper, the spring and the earth is exchanged only with each other not outside from the system. Since the earth and spring act on the jumper, the system is isolated.

Conclusion:

Therefore, the system the bungee jumper, the spring and the earth is isolated.

(d)

Expert Solution
Check Mark
To determine

The spring constant of the bungee cord.

Answer to Problem 14P

The spring constant of the bungee cord is 73.38N/m2.

Explanation of Solution

Write an expression of the law of conservation of the energy for the bungee jumper

    mgh=12kh12

Here, m is the mass of jumper, g is the acceleration due to gravity, h is the total height, k is the spring constant and h1 is the height of simple harmonic oscillation.

Substitute 65.0kg for m, 9.8m/s2 for g, 36.0m for h and 25.0m for h1 in above equation to find k.

    (65.0kg)(9.8m/s2)(36.0m)=12k(25.0m)2k=2(65.0kg)(9.8m/s2)(36.0m)(25.0m)2=73.38N/m2

Conclusion:

Therefore, the spring constant of the bungee cord is 73.38N/m2.

(e)

Expert Solution
Check Mark
To determine

The location of the equilibrium point where the spring force balances the gravitational force exerted on the jumper.

Answer to Problem 14P

The location of the equilibrium point where the spring force balances the gravitational force exerted on the jumper is 16.32m.

Explanation of Solution

The equation for the equilibrium point from the lorded is,

    kx=mgx=mgk

Here, x is the position of the spring.

The equilibrium point is calculated as,

    y=L+x

Here, L is the stretched length of the chord.

Substitute mgk for x in above expression.

    y=L+(mgk)

Substitute 65.0kg for m, 9.8m/s2 for g, 73.38N/m2 for k and 11.0m for L in above equation to find y.

    y=(11.0m)+(65.0kg)(9.8m/s2)(73.38N/m2)=19.68m

The amplitude of the motion is,

    A=hy

Substitute 36.0m for h and 19.68m for y in above equation to find A.

    A=(36.0m)(19.68m)=16.32m

Conclusion:

Therefore, the location of the equilibrium point where the spring force balances the gravitational force exerted on the jumper is 16.32m.

(f)

Expert Solution
Check Mark
To determine

The angular frequency of the oscillation.

Answer to Problem 14P

The angular frequency of the oscillation is 1.06rad/s.

Explanation of Solution

The formula to calculate angular frequency of the oscillation is,

    ω=km

Substitute 65.0kg for m and 73.38N/m2 for k in above equation to find ω.

    ω=73.38N/m265.0kg=1.06rad/s

Conclusion:

Therefore, the angular frequency of the oscillation is 1.06rad/s.

(g)

Expert Solution
Check Mark
To determine

The time interval required for the cord to stretched by 25.0m.

Answer to Problem 14P

The time interval required for the cord to stretched by 25.0m is 2.0s.

Explanation of Solution

The expression for the position of a particle in simple harmonic motion is,

    x=Acos(ωt1)

Substitute mgk for x in above expression.

    mgk=Acos(ωt1)

Substitute 65.0kg for m, 9.8m/s2 for g, 73.38N/m2 for k, 16.32m for A and 1.06rad/s for ω in above equation to find t1.

    (65.0kg)(9.8m/s2)(73.38N/m2)=(16.32m)cos((1.06rad/s)t1)8.68m(16.32m)=cos((1.06rad/s)t1)0.53=cos((1.06rad/s)t1)t1=2.0s

Conclusion:

Therefore, the time interval required for the cord to stretched by 25.0m is 2.0s.

(h)

Expert Solution
Check Mark
To determine

The total time interval for the entire 36.0m drop.

Answer to Problem 14P

The total time interval for the entire 36.0m drop is 3.49s.

Explanation of Solution

The total time interval for the entire 36.0m drop is,

    T=t+t1

Substitute 1.49s for t and 2.0s for t1 in above equation to find T.

    T=1.49s+2.0s=3.49s

Conclusion:

Therefore, the total time interval for the entire 36.0m drop is 3.49s.

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Chapter 15 Solutions

Physics for Scientists and Engineers

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