/*! 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 83 In the dangerous "sport" of bung... [FREE SOLUTION] | 91Ó°ÊÓ

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In the dangerous "sport" of bungee jumping, a daring student jumps from a hot- air balloon with a specially designed elastic cord attached to his waist. The unstretched length of the cord is \(25.0 \mathrm{~m}\), the student weighs \(700 \mathrm{~N}\), and the balloon is \(36.0 \mathrm{~m}\) above the surface of a river below. Calculate the required force constant of the cord if the student is to stop safely \(4.00 \mathrm{~m}\) above the river.

Short Answer

Expert verified
The required force constant of the cord for the student to land safely is approximately \(151.6 N/m\).

Step by step solution

01

Identify Knowns and Unknowns

In this problem, we know that the weight (force due to gravity) of the student is \(700 \mathrm{~N}\), the initial jumping height is \(36.0 \mathrm{~m}\), the unstretched length of the elastic cord is \(25.0 \mathrm{~m}\), and the student should stop \(4.00 \mathrm{~m}\) above the river. We aim to find the spring constant of the cord.
02

Express Energy at the Starting Point and Lowest Point

At the start, all energy is in the form of gravitational potential energy (PE). At the lowest point (when the cord is fully stretched and the student is \(4.00 \mathrm{~m}\) above the river), all energy is transferred into the potential energy of the stretched cord (since ideally no energy is lost during the fall). Therefore, the gravitational potential energy at the start should equal the elastic potential energy at the lowest point. This gives us: \[PE_{gravity} = PE_{spring}\].
03

Substitute Knowns and Solve for Unknown

Substituting the formulas for gravitational and elastic potential energy in the equation gives: \[mgΔh = 0.5kx^2\]\(m\) is the mass of the student, \(g\) is the acceleration due to gravity (approximately \(9.8 \mathrm{m/s^2}\)), Δh is the height the student falls before the cord starts to stretch, \(k\) is the spring constant, and \(x\) is the total stretch of the cord. Note that Δh is the falling height before the cord starts to stretch (which is the initial height minus the cord length) and \(x\) is the total stretch when the student stops above the river (which is the falling height minus the minimum distance above the river).The force due to gravity \(F = mg\) and we know \(F = 700N\), so we can substitute \(mg\) with \(700N\). By doing so and further rearranging for \(k\), we get \(k = \frac{2 \times 700N \times Δh}{x^2}\).
04

Substitute Values and Calculate

Finally, we substitute the known values back into our equation to find the spring constant. Here, Δh = \(36m - 25m = 11m\) and \(x = 36m - 4m = 32m\). Substituting these values gives \(k = \frac{2 \times 700N \times 11m}{32m^2} \approx 151.6 N/m\). So, the required force constant of the cord is approximately \(151.6 N/m\).

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Key Concepts

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

Gravitational Potential Energy
When considering the energy utilized in bungee jumping, a significant factor is gravitational potential energy (GPE). GPE is the energy an object possesses due to its position in a gravitational field. In the context of our bungee jumping scenario, GPE is highest at the starting point, where the student initially jumps from the balloon.

The formula for calculating gravitational potential energy is: \[ GPE = mgh \]where
  • \( m \) is the object's mass,
  • \( g \) is the acceleration due to gravity—commonly approximated as \(9.8 \mathrm{m/s^2}\) on Earth,
  • \( h \) is the height above a reference point—in this case, the distance from the balloon to the river.
In our exercise, the student's GPE at the start measures the potential for the student to do work on the bungee cord. As the student falls, this energy is converted to kinetic energy and, eventually, to elastic potential energy stored in the stretched bungee cord.
Spring Constant
The spring constant, symbolized by \( k \), is a measure of the stiffness of a spring. It connects the force exerted by the spring to the displacement caused by that force, as defined by Hooke's Law, which states:\[ F = -kx \]
  • \( F \) is the force exerted by the spring,
  • \( k \) is the spring constant,
  • \( x \) is the displacement from the spring's equilibrium position.
In the case of bungee jumping, the elastic cord acts like a spring. The spring constant defines how 'tough' the cord needs to be to stop the student at a safe distance above the water. If \( k \) is too low, the cord will stretch too much, risking contact with the river. If \( k \) is too high, the cord may not stretch sufficiently, causing a high force on the jumper at the rebound which can be dangerous.
Conservation of Mechanical Energy
Conservation of mechanical energy is a fundamental principle in physics which states that in the absence of non-conservative forces (like air resistance or friction), the total mechanical energy of a system remains constant. Mechanical energy is the sum of kinetic and potential energy in a system. For our bungee jumper, this principle dictates that the initial gravitational potential energy will be transformed into elastic potential energy without loss, when air resistance is negligible.

The equation representing this conservation is:\[ PE_{initial} + KE_{initial} = PE_{final} + KE_{final} \]where
  • \( PE_{initial} \) and \( KE_{initial} \) are the potential and kinetic energy at the beginning, and
  • \( PE_{final} \) and \( KE_{final} \) are the potential and kinetic energy at the lowest point of descent.
In the ideal scenario of our exercise, the student’s initial GPE is converted entirely into the elastic potential energy of the cord, which is formulated by assuming that the kinetic energy at the lowest point of descent is zero (the instant before the student starts to ascend back up). This allows us to set the energies equal and solve for the unknown spring constant.

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Most popular questions from this chapter

A \(0.60-\mathrm{kg}\) particle has a speed of \(2.0 \mathrm{~m} / \mathrm{s}\) at point \(A\) and a kinetic energy of 7.5 J at point \(B\). What is (a) its kinetic energy at \(A\) ? (b) Its speed at point \(B\) ? (c) The total work done on the particle as it moves from \(A\) to \(B\) ?

A light spring with force constant \(3.85 \mathrm{~N} / \mathrm{m}\) is compressed by \(8.00 \mathrm{~cm}\) as it is held between a \(0.250-\mathrm{kg}\) block on the left and a \(0.500\)-kg block on the right, both resting on a horizontal surface. The spring exerts a force on each block, tending to push them apart. The blocks are simultaneously released from rest. Find the acceleration with which each block starts to move, given that the coefficient of kinetic friction between each block and the surface is (a) 0 , (b) \(0.100\), and (c) \(0.462\).

A \(50.0-\mathrm{kg}\) projectile is fired at an angle of \(30.0^{\circ}\) above the horizontal with an initial speed of \(1.20 \times 10^{2} \mathrm{~m} / \mathrm{s}\) from the top of a cliff \(142 \mathrm{~m}\) above level ground, where the ground is taken to be \(y=0\). (a) What is the initial total mechanical energy of the projectile? (b) Suppose the projectile is traveling \(85.0 \mathrm{~m} / \mathrm{s}\) at its maximum height of \(y=427 \mathrm{~m}\). How much work has been done on the projectile by air friction? (c) What is the speed of the projectile immediately before it hits the ground if air friction does one and a half times as much work on the projectile when it is going down as it did when it was going up?

\(\mathrm{S}\) A projectile of mass \(m\) is fired horizontally with an initial speed of \(v_{0}\) from a height of \(h\) above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of \(m, v_{0}, h\), and \(g\) : (a) the work done by the force of gravity on the projectile, (b) the change in kinetic energy of the projectile since it was fired, and (c) the final kinetic energy of the projectile. (d) Are any of the answers changed if the initial angle is changed?

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