/*! 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 59 A stone with a weight of \(5.29 ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A stone with a weight of \(5.29 \mathrm{~N}\) is launched vertically from ground level with an initial speed of \(20.0 \mathrm{~m} / \mathrm{s}\), and the air drag on it is \(0.265 \mathrm{~N}\) throughout the flight. What are (a) the maximum height reached by the stone and (b) its speed just before it hits the ground?

Short Answer

Expert verified
(a) The maximum height is 21.48 meters. (b) The speed just before impact is 20 m/s.

Step by step solution

01

Determine the Stone's Mass

To find the maximum height, we first need to determine the stone's mass using its weight. The weight of the stone is given as \(5.29\, \, \mathrm{N}\), which is the force due to gravity. Using the formula: \( F = mg \), where \( g \approx 9.8 \, \mathrm{m/s^2} \), we calculate the mass \( m \): \( m = \frac{5.29\, \mathrm{N}}{9.8\, \mathrm{m/s^2}} \approx 0.54 \mathrm{~kg} \).
02

Calculate Net Force and Net Acceleration

The air drag is given as \(0.265\, \mathrm{N}\). To find the net force acting on the object while it is rising, subtract the drag force from the weight since they oppose the upward movement: \( F_{\text{net}} = -(5.29 \, \mathrm{N}) + 0.265 \, \mathrm{N} = -5.025 \, \mathrm{N} \). Using \( F = ma \), solve for \( a \): \( a = \frac{-5.025 \mathrm{~N}}{0.54 \mathrm{~kg}} \approx -9.31 \mathrm{~m/s^2} \).
03

Find the Maximum Height

Apply kinematic equations to find the maximum height. The initial velocity \( v_0 = 20.0 \mathrm{~m/s} \) and at the max height, the final velocity \( v = 0 \mathrm{~m/s} \). Use the equation \( v^2 = v_0^2 + 2a s \), solving for \( s \): \( 0 = (20)^2 + 2(-9.31)s \), \( s = \frac{400}{18.62} \approx 21.48 \mathrm{~m} \).
04

Calculate Speed Just Before Impact

Upon reaching the ground, we assume it falls back with the same net acceleration but in the opposite direction. Using \( v^2 = u^2 + 2as \) where \( u = 0 \) initially after reach max height, \( a = 9.31\, \mathrm{m/s^2} \), and \( s = 21.48 \mathrm{~m} \): \( v^2 = 0 + 2(9.31)(21.48) \), \( v = \sqrt{399.53} \approx 20 \mathrm{~m/s} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Kinematics
In the realm of physics, kinematics stands as the branch that focuses on the description of motion without considering the forces that cause it. In projectile motion, such as when a stone is thrown vertically upwards, we examine the initial velocity, maximum height, and time of flight. These aspects are crucial in predicting the trajectory and impact of an object.
When we talk about vertical projectile motion, the initial velocity and acceleration due to gravity play pivotal roles. The stone in our problem is launched with an initial speed of 20.0 m/s, and throughout its journey upwards, it fights against both gravity and air drag. The kinematic equations lend themselves well to finding the unknowns here.
  • Initial velocity, noted as \( v_0 \), starts at 20.0 m/s.
  • Acceleration due to gravity, \( g \), which is 9.8 m/s², constantly pulls the stone downwards.
  • At its maximum height, the stone's velocity (\( v \)) becomes 0 m/s.
  • The kinematic equation \( v^2 = v_0^2 + 2a s \) helps determine distances involved without referencing time directly.
Understanding the interaction of these elements helps us calculate the maximum height the stone reaches, which is a remarkable 21.48 meters in this exercise.
Forces and Acceleration
Forces are at the heart of understanding motion dynamics. Here, two main forces affect the stone's flight: gravitational force and air resistance. The gravitational force is straightforward, acting downward, while air resistance acts against the stone's motion.
To analyze these forces, we quantify the stone's weight as 5.29 N. Using the formula \( F = mg \), we can determine that the stone's mass is approximately 0.54 kg. This is essential since mass links force and acceleration.
The presence of air drag (0.265 N), opposing the stone's ascent, modifies the net force acting on it:
  • Net force \( F_{\text{net}} \) is calculated as the difference between the gravitational force and air drag.
  • Using Newton's second law \( F = ma \), we derive the net acceleration as \(-9.31 \text{ m/s}^2 \).
These calculations explain why the stone doesn’t rise indefinitely—it slows down due to air drag and gravity, reverses direction, and accelerates again towards the ground. By understanding forces and accelerations, we link the physical push and pull acting on bodies to their resultant movements.
Energy Conservation
Energy plays a fundamental role in understanding the motion of projectiles. In an ideal scenario without air resistance, we could conserve mechanical energy to analyze motion. However, in this exercise, energy conservation considerations become slightly more complex due to the air drag.
Despite of these non-conservative forces like air drag, we observe conservation principles when calculating speed before impact. When a stone is thrown upwards from the ground, it initially possesses kinetic energy due to its velocity:
  • Kinetic energy at launch is quantified by \(\frac{1}{2}mv_0^2\).
  • As it ascends, kinetic energy converts to gravitational potential energy \(mgh\) until the stone comes momentarily to a stop at the apex.
Upon descent, potential energy shifts back to kinetic energy. Accounting for the forces present, the stone regains a speed of approximately 20 m/s before hitting the ground. This example powerfully showcases how energy transformations govern projectile motion, all while accounting for opposing forces that slightly alter the calculation from an ideal scenerio.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Each second, \(1200 \mathrm{~m}^{3}\) of water passes over a waterfall \(100 \mathrm{~m}\) high. Three-fourths of the kinetic energy gained by the water in falling is transferred to electrical energy by a hydroelectric generator. At what rate does the generator produce electrical energy? (The mass of \(1 \mathrm{~m}^{3}\) of water is \(1000 \mathrm{~kg}\).)

A 60 kg skier starts from rest at height H 20 m above the end of a ski-jump ramp (Fig. 8-37) and leaves the ramp at angle u 28. Neglect the effects of air resistance and assume the ramp is frictionless. (a) What is the maximum height h of his jump above the end of the ramp? (b) If he increased his weight by putting on a backpack, would h then be greater, less, or the same?

From the edge of a cliff, a \(0.55 \mathrm{~kg}\) projectile is launched with an initial kinetic energy of \(1550 \mathrm{~J}\). The projectile's maximum upward displacement from the launch point is \(+140 \mathrm{~m}\). What are the (a) horizontal and (b) vertical components of its launch velocity? (c) At the instant the vertical component of its velocity is \(65 \mathrm{~m} / \mathrm{s}\), what is its vertical displacement from the launch point?

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 s interval? (c) What is the instantaneous power at the end of the 30 s interval, assuming that the acceleration is constant?

A pendulum consists of a \(2.0 \mathrm{~kg}\) stone swinging on a 4.0 \(\mathrm{m}\) string of negligible mass. The stone has a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) when it passes its lowest point. (a) What is the speed when the string is at \(60^{\circ}\) to the vertical? (b) What is the greatest angle with the vertical that the string will reach during the stone's motion? (c) If the potential energy of the pendulum-Earth system is taken to be zero at the stone's lowest point, what is the total mechanical energy of the system?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.