/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 70 It's your birthday, and to celeb... [FREE SOLUTION] | 91Ó°ÊÓ

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It's your birthday, and to celebrate you're going to make your first bungee jump. You stand on a bridge 100 m above a raging river and attach a 30 -m-long bungee cord to your harness. A bungee cord, for practical purposes, is just a long spring, and this cord has a spring constant of 40 N/m. Assume that your mass is 80 kg. After a long hesitation, you dive off the bridge. How far are you above the water when the cord reaches its maximum elongation?

Short Answer

Expert verified
The maximum elongation of the bungee cord will be found by solving the equation for x and then subtracting this value from the initial height.

Step by step solution

01

Calculate Initial Potential Energy

When the jumper is on the platform, all of the energy is potential energy. Calculate the initial potential energy using the formula: PE = m*g*h, where m is the mass (80 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (100 m). So, PE = 80*9.8*100 = 78400 J. This is the total energy at the start.
02

Calculate Maximum Elastic Potential Energy

When the bungee cord is fully extended, all the energy will have transferred to the elastic potential energy in the bungee cord. Calculate the elastic potential energy using the formula: EPE = 0.5*k*x^2, where k is the spring constant (40 N/m) and x is the unknown maximum extension of the cord. At this point, all kinetic energy has been transferred to this elastic potential energy, so EPE = PE.
03

Solve for Maximum Extension

To find the unknown maximum elongation (x), equate the initial potential energy to the elastic potential energy, 78400 = 0.5*40*x^2. Solve this to find the value of x. Once you have x, subtract it from the initial height of the jump to determine how far above the water you would be when the cord has reached maximum elongation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Energy
The concept of the 'conservation of energy' states that energy cannot be created or destroyed, only transformed from one form to another. In the context of bungee jumping, it's the key principle that allows us to predict how the system will behave. The jumper starts with a certain amount of gravitational potential energy at the top of the bridge. As they fall, this energy is converted into kinetic energy - the energy of motion. At the maximum elongation of the bungee cord, the kinetic energy is then converted into elastic potential energy within the cord itself. This energy transformation continues to cycle until external forces such as air resistance and friction within the cord dampen the motion to a halt. When calculating these energies, if our system is assumed to be isolated (ignoring air resistance for simplicity), the total energy remains constant throughout the jump. This is crucial for solving bungee jumping physics problems accurately.
Elastic Potential Energy
Elastic potential energy is the energy stored in elastic materials as the result of their stretching or compressing. In a bungee jumping scenario, the bungee cord can be likened to a spring. The cord stores energy as it stretches, similar to how a compressed spring stores energy. This energy can be calculated using the formula:
\( EPE = \frac{1}{2}kx^2 \),
where \( EPE \) represents the elastic potential energy, \( k \) is the spring constant, indicating the stiffness of the cord, and \( x \) is the amount of stretch or elongation of the cord. Understanding how elastic potential energy works is essential for determining how the energy of the jumper's fall is stored and released by the bungee cord.
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field, like that of the Earth. The higher the object is lifted, the greater the stored energy, because there's more distance to fall. For an object of mass \( m \), at a height \( h \) above the ground in the Earth's gravitational field (with gravity \( g = 9.8 \) m/s^2), the GPE is given by the formula:
\( GPE = mgh \).
In the bungee jumping problem, we calculate the initial GPE at the moment before the jumper leaves the bridge. This calculated GPE represents the total mechanical energy the system initially has, which will eventually convert into elastic potential energy once the bungee cord is fully stretched.
Spring Constant
The spring constant, denoted as \( k \), is a measure of a spring's (or in bungee jumping, the bungee cord's) stiffness. Mathematically, it's the ratio of the force affecting the spring to the displacement caused by it. A higher spring constant means a stiffer spring, which requires more force to stretch or compress it. The SI unit for the spring constant is newtons per meter (N/m). In the provided bungee jumping problem, the bungee cord's spring constant is 40 N/m. This value is used to calculate the maximum elastic potential energy in the cord and understand how much the cord can stretch in response to the force exerted by the jumper's mass as they fall.

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Most popular questions from this chapter

A \(50 \mathrm{g}\) ice cube can slide without friction up and down a \(30^{\circ}\) slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring \(10 \mathrm{cm} .\) The spring constant is 25 N/m. When the ice cube is released, what distance will it travel up the slope before reversing direction?

A \(100 \mathrm{g}\) ball moving to the right at \(4.0 \mathrm{m} / \mathrm{s}\) catches up and collides with a \(400 \mathrm{g}\) ball that is moving to the right at \(1.0 \mathrm{m} / \mathrm{s} .\) If the collision is perfectly elastic, what are the speed and direction of each ball after the collision?

A vertical spring with \(k=490 \mathrm{N} / \mathrm{m}\) is standing on the ground. You are holding a \(5.0 \mathrm{kg}\) block just above the spring, not quite touching it. a. How far does the spring compress if you let go of the block suddenly? b. How far does the spring compress if you slowly lower the block to the point where you can remove your hand without disturbing it? c. Why are your two answers different?

Protons and neutrons (together called nucleons) are held together in the nucleus of an atom by a force called the strong force. At very small separations, the strong force between two nucleons is larger than the repulsive electrical force between two protons - hence its name. But the strong force quickly weakens as the distance between the protons increases. A well-established model for the potential energy of two nucleons interacting via the strong force is $$U=U_{0}\left[1-e^{-x / x_{0}}\right]$$ where \(x\) is the distance between the centers of the two nucleons, \(x_{0}\) is a constant having the value \(x_{0}=2.0 \times 10^{-15} \mathrm{m},\) and \(U_{0}=6.0 \times 10^{-11} \mathrm{J}\) a. Calculate and draw an accurate potential-energy curve from \(x=0 \mathrm{m}\) to \(x=10 \times 10^{-15} \mathrm{m} .\) Either calculate about 10 points by hand or use computer software. b. Quantum effects are essential for a proper understanding of how nucleons behave. Nonetheless, let us innocently consider two neutrons as if they were small, hard, electrically neutral spheres of mass \(1.67 \times 10^{-27} \mathrm{kg}\) and diameter \(1.0 \times 10^{-15} \mathrm{m}\) (We will consider neutrons rather than protons so as to avoid complications from the electric forces between protons.) You are going to hold two neutrons \(5.0 \times 10^{-15} \mathrm{m}\) apart, measured between their centers, then release them. Draw the total energy line for this situation on your diagram of part a. c. What is the speed of each neutron as they crash together? Keep in mind that both neutrons are moving.

A 10 kg box slides 4.0 m down the frictionless ramp shown then collides with a spring whose spring constant is \(250 \mathrm{N} / \mathrm{m}\) a. What is the maximum com- pression of the spring? b. At what compression of the spring does the box have its maximum speed?

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