/*! 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 111 A \(9.40 \mathrm{~kg}\) projecti... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(9.40 \mathrm{~kg}\) projectile is fired vertically upward. Air drag decreases the mechanical energy of the projectile-Earth system by \(68.0 \mathrm{~kJ}\) during the projectile's ascent. How much higher would the projectile have gone were air drag negligible?

Short Answer

Expert verified
The projectile would have gone 737.4 meters higher without air drag.

Step by step solution

01

Understand the Problem

First, identify the given information and what is being asked. We know that a projectile with a mass of \(9.40 \text{ kg}\) loses \(68.0 \text{ kJ}\) of mechanical energy due to air drag. We need to find out how much additional height the projectile would have achieved if there was no air drag.
02

Convert Energy Units

Convert the energy lost from kilojoules to joules for easier calculation.\[68.0 \text{ kJ} = 68,000 \text{ J}\]This is the additional mechanical energy that would have been available for the projectile to ascend higher.
03

Establish the Formula

Use the gravitational potential energy formula, \(U = mgh\), where \(U\) is the potential energy, \(m\) is the mass, \(g\) is the acceleration due to gravity (approximately \(9.81 \text{ m/s}^2\)), and \(h\) is the height.
04

Rearrange the Formula

Rearrange the formula \(U = mgh\) to solve for the height \(h\):\[h = \frac{U}{mg}\]This formula will be used to find the additional height the projectile would attain.
05

Calculate the Additional Height

Substitute the known values into the equation:\[ h = \frac{68,000 \text{ J}}{9.40 \text{ kg} \times 9.81 \text{ m/s}^2}= \frac{68,000}{92.214} \text{ m} = 737.4 \text{ m}\]Thus, the projectile would have risen an additional 737.4 meters if there had been no air drag.

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. For objects near the Earth's surface, this energy depends primarily on the object's mass, the height it is above the ground, and the gravitational acceleration. The formula for gravitational potential energy is expressed as:\[ U = mgh \]where:
  • \( U \) is the gravitational potential energy (measured in joules),
  • \( m \) is the mass of the object (in kilograms),
  • \( g \) is the acceleration due to gravity, roughly \( 9.81 \, \text{m/s}^2 \),
  • \( h \) is the height above the ground (in meters).
In the given problem, the projectile has gravitational potential energy which increases as it ascends into the air. The higher the projectile goes, the more potential energy it gains. However, air resistance affects how much of this energy can be converted completely during the ascent. Understanding this concept helps explain why energy losses due to external forces like air friction can affect the height achieved by the projectile.
Air Resistance
Air resistance, often known as drag, is a force that opposes an object's motion through air. It acts in the opposite direction to the object's velocity and increases with the speed of the object. This force is a crucial factor in real-world physics problems because it causes a loss of mechanical energy that would otherwise contribute to an object's desired trajectory or height. In the scenario of the vertically-fired projectile:
  • Air resistance converts some of the projectile's kinetic energy into other forms of energy like heat.
  • This results in a reduction of the mechanical energy available for further ascent.
  • Air resistance makes it more challenging for the projectile to reach a higher altitude.
In this exercise, we know that the air resistance led to a 68,000 joule (or 68 kJ) loss in energy, preventing the projectile from attaining its maximum possible height. This highlights why understanding and calculating the effect of air resistance is vital when analyzing projectile motions.
Mechanical Energy Loss
Mechanical energy loss in a system is the reduction of the total mechanical energy due to non-conservative forces like friction or air resistance. In the context of this exercise, mechanical energy loss occurs as the projectile ascends and encounters air drag. Here are the key points to understand about mechanical energy loss:
  • Energy can be converted from one form to another but not all energy stays in the form of mechanical energy when non-conservative forces are at play.
  • Mechanical energy loss due to air resistance results in less energy being available for gravitational potential energy as the projectile ascends.
  • This loss of energy manifests as a reduction in the maximum height that the projectile can achieve.
By calculating the potential increase in height without air resistance, as done in the step-by-step solution, it becomes clear how significant the impact of mechanical energy loss can be on projectile motion. It's essential to take these losses into account when performing accurate physical predictions.

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Most popular questions from this chapter

A \(1500 \mathrm{~kg}\) car starts from rest on a horizontal road and gains a speed of \(72 \mathrm{~km} / \mathrm{h}\) in \(30 \mathrm{~s}\). (a) What is its kinetic energy at the end of the \(30 \mathrm{~s}\) ? (b) What is the average power required of the car during the \(30 \mathrm{~s}\) interval? (c) What is the instantaneous power at the end of the 30 s interval, assuming that the acceleration is constant?

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The potential energy of a diatomic molecule (a two-atom system like \(\mathrm{H}_{2}\) or \(\mathrm{O}_{2}\) ) is given by $$U=\frac{A}{r^{12}}-\frac{B}{r^{6}}$$ where \(r\) is the separation of the two atoms of the molecule and \(A\) and \(B\) are positive constants. This potential energy is associated with the force that binds the two atoms together. (a) Find the equilibrium separation - that is, the distance between the atoms at which the force on each atom is zero. Is the force repulsive (the atoms are pushed apart) or attractive (they are pulled together) if their separation is (b) smaller and (c) larger than the equilibrium separation?

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