/*! 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 skier starts from rest at the top of a hill that is inclined \(10.5^{\circ}\) with respect to the horizontal. The hillside is \(200 \mathrm{~m}\) long, and the coefficient of friction between snow and skis is \(0.0750\). At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier glide along the horizontal portion of the snow before coming to rest?

A child's pogo stick (Fig. P5.81) stores energy in a spring \((k=\) \(\left.2.50 \times 10^{4} \mathrm{~N} / \mathrm{m}\right)\). At position (A) \(\left(x_{1}=-0.100 \mathrm{~m}\right)\), the spring compression is a maximum and the child is momentarily at rest. At position (B) \((x=0)\), the spring is relaxed and the child is moving upward. At position (C), the child is again momentarily at rest at the top of the jump. Assuming that Figure p \(5.81\) the combined mass of child and pogo stick is \(25.0 \mathrm{~kg}\), (a) calculate the total energy of the system if both potential energies are zero at \(x=0\), (b) determine \(x_{2}\), (c) calculate the speed of the child at \(x=0\), (d) determine the value of \(x\) for which the kinetic energy of the system is a maximum, and (c) obtain the child's maximum upward speed.

A \(65.0-\mathrm{kg}\) runner has a speed of \(5.20 \mathrm{~m} / \mathrm{s}\) at one instant during a long-distance event. (a) What is the runner's kinetic energy at this instant? (b) If he doubles his speed to reach the finish line, by what factor does his kinetic energy change?

GP A horizontal spring attached to a wall has a force constant of \(850 \mathrm{~N} / \mathrm{m}\). A block of mass \(1.00 \mathrm{~kg}\) is attached to the spring and oscillates freely on a horizontal, frictionless surface as in Figure \(5.20\). The initial goal of this problem is to find the velocity at the equilibrium point after the block is released. (a) What objects constitute the system, and through what forces do they interact? (b) What are the two points of interest? (c) Find the energy stored in the spring when the mass is stretched \(6.00 \mathrm{~cm}\) from equilibrium and again when the mass passes through equilibrium after being released from rest. (d) Write the conservation of energy equation for this situation and solve it for the speed of the mass as it passes equilibrium. Substitute to obtain a numerical value. (c) What is the speed at the halfway point? Why isn't it half the speed at equilibrium?

A large cruise ship of mass \(6.50 \times 10^{7} \mathrm{~kg}\) has a speed of \(12.0 \mathrm{~m} / \mathrm{s}\) at some instant. (a) What is the ship's kinetic cnergy at this time? (b) How much work is required to stop it? (c) What is the magnitude of the constant force required to stop it as it undergoes a displacement of \(2.50 \mathrm{~km}\) ?

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