/*! 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 116 A \(68 \mathrm{~kg}\) sky diver ... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 116

A \(68 \mathrm{~kg}\) sky diver falls at a constant terminal speed of \(59 \mathrm{~m} / \mathrm{s}\). (a) At what rate is the gravitational potential energy of the Earth-sky diver system being reduced? (b) At what rate is the system's mechanical energy being reduced?

Short Answer

Expert verified
(a) 39,268 W; (b) 39,268 W

Step by step solution

01

Understand the Problem

We need to find two different rates for the sky diver system. First, we will calculate the rate at which gravitational potential energy is reduced, then the rate at which mechanical energy is reduced as the sky diver falls at terminal speed.
02

Determine the Rate of Change of Gravitational Potential Energy

Gravitational potential energy (U) is given by the formula: \[ U = mgh \]where \( m \) is mass, \( g \) is acceleration due to gravity (\( 9.81 \, \mathrm{m/s^2} \)), and \( h \) is height. The rate of change of potential energy with respect to time is given by:\[ \frac{dU}{dt} = \frac{d}{dt}(mgh) = mg \frac{dh}{dt} \]Since \( \frac{dh}{dt} \) is the velocity, at terminal speed, it is \( 59 \, \mathrm{m/s} \). Thus,\[ \frac{dU}{dt} = 68 \, \mathrm{kg} \times 9.81 \, \mathrm{m/s^2} \times 59 \, \mathrm{m/s} = 39,267.72 \, \mathrm{W} \]This means the gravitational potential energy is being reduced at a rate of 39,268 W.
03

Determine the Rate of Change of Mechanical Energy

At terminal speed, the kinetic energy is constant, implying that the mechanical energy change is only due to the change in potential energy. Thus, the rate of change of mechanical energy is equal to the rate of change of potential energy. So:\[ \text{Rate of change of mechanical energy} = 39,268 \, \mathrm{W} \]

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Key Concepts

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

Mechanical Energy
Mechanical energy is the sum of two types of energy: kinetic energy and potential energy. Kinetic energy refers to the energy that an object possesses because of its motion. Potential energy, on the other hand, is the energy stored due to an object's position or configuration. In the context of the skydiver, both of these energies play a role.
  • Kinetic Energy: Given by the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity. For a skydiver at terminal velocity, kinetic energy remains constant because the velocity does not change.
  • Potential Energy: Expressed as \( PE = mgh \), where \( h \) represents height. As the skydiver falls, the height decreases, which leads to a reduction in potential energy over time.
Thus, for the skydiver, as she maintains a constant terminal velocity, her mechanical energy changes primarily due to the decrease in gravitational potential energy, since kinetic energy remains stable.
Terminal Velocity
Terminal velocity is a key concept in the physics of falling objects. It occurs when an object falling through a fluid (like air) reaches a speed where the resistance, or drag force, from the fluid equals the gravitational force pulling the object down.
  • At terminal velocity, the net force on the object is zero, hence it no longer accelerates and falls at a constant speed.
  • For a skydiver, this means she's moving downward at a steady rate, which for this problem is \(59 \, \mathrm{m/s}\).
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Terminal velocity is crucial because it simplifies the calculations of energy changes. When a body falls at a terminal velocity, we can assume constant mechanical energy unless other external forces act on it.
Energy Conservation
In physics, the principle of conservation of energy is paramount. This principle states that energy in a closed system is conserved; it can neither be created nor destroyed, only transformed from one form to another.
  • As the skydiver descends at terminal speed, her potential energy is continuously transformed into kinetic energy.
  • However, since she is already at terminal velocity, no net increase in kinetic energy occurs as the majority of this energy transfers as thermal energy due to air resistance.
  • The reduction in potential energy directly translates to energy dissipated into the atmosphere, emphasizing that total energy within the system remains constant over time.
Energy conservation helps explain why the rate of potential energy reduction matches the rate of mechanical energy reduction in the skydiver's descent.
Physics Problem Solving
Solving physics problems often requires a strategic approach. By breaking down the problem into manageable steps, as shown in the exercise, you can systematically address each aspect of the task.
  • Identify Knowns and Unknowns: Determine what information you have and what you need to find. In this case, key variables included mass, gravitational acceleration, and terminal velocity.
  • Apply Relevant Principles: Use physics laws and formulas relevant to the problem. Here, using formulas for potential energy and understanding terminal velocity were crucial.
  • Check Your Work: Ensure that your solution is logical, and units are consistent. This confirms that the answers make sense within the physical context.
When tackling physics problems, patience and attention to detail are important. Understanding the underlying concepts through a step-by-step approach helps develop problem-solving skills efficiently.

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Most popular questions from this chapter

A \(0.50 \mathrm{~kg}\) banana is thrown directly upward with an initial speed of \(4.00 \mathrm{~m} / \mathrm{s}\) and reaches a maximum height of \(0.80 \mathrm{~m}\). What change does air drag cause in the mechanical energy of the banana-Earth system during the ascent?

A conservative force \(\vec{F}=(6.0 x-12) \hat{\mathrm{i}} \mathrm{N}\) where \(x\) is in meters, acts on a particle moving along an \(x\) axis. The potential energy \(U\) associated with this force is assigned a value of \(27 \mathrm{~J}\) at \(x=0\). (a) Write an expression for \(U\) as a function of \(x\), with \(U\) in joules and \(x\) in meters. (b) What is the maximum positive potential energy? At what (c) negative value and (d) positive value of \(x\) is the potential energy equal to zero?

A \(5.0 \mathrm{~kg}\) block is projected at \(5.0 \mathrm{~m} / \mathrm{s}\) up a plane that is inclined at \(30^{\circ}\) with the horizontal. How far up along the plane does the block go (a) if the plane is frictionless and (b) if the coefficient of kinetic friction between the block and the plane is \(0.40 ?\) (c) In the latter case, what is the increase in thermal energy of block and plane during the block's ascent? (d) If the block then slides back down against the frictional force, what is the block's speed when it reaches the original projection point?

A \(0.42 \mathrm{~kg}\) shuffleboard disk is initially at rest when a player uses a cue to increase its speed to \(4.2 \mathrm{~m} / \mathrm{s}\) at constant acceleration. The acceleration takes place over a \(2.0 \mathrm{~m}\) distance, at the end of which the cue loses contact with the disk. Then the disk slides an additional \(12 \mathrm{~m}\) before stopping. Assume that the shuffleboard court is level and that the force of friction on the disk is constant. What is the increase in the thermal energy of the disk-court system (a) for that additional \(12 \mathrm{~m}\) and \((\mathrm{b})\) for the entire \(14 \mathrm{~m}\) distance? (c) How much work is done on the disk by the cue?

A worker pushed a \(27 \mathrm{~kg}\) block \(9.2 \mathrm{~m}\) along a level floor at constant speed with a force directed \(32^{\circ}\) below the horizontal. If the coefficient of kinetic friction between block and floor was \(0.20\), what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block- floor system?
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