/*! 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 41 A 47.0-g golf ball is driven fro... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 41

A 47.0-g golf ball is driven from the tee with an initial speed of \(52.0 \mathrm{m} / \mathrm{s}\) and rises to a height of \(24.6 \mathrm{m}\). (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is \(8.0 \mathrm{m}\) below its highest point?

Short Answer

Expert verified
(a) KE at highest point = 52.36 J; (b) Speed 8 m below = 48.81 m/s.

Step by step solution

01

Understand the concepts involved

We need to use the concepts of kinetic energy (KE) and potential energy (PE). The law of conservation of mechanical energy states that the total mechanical energy (sum of KE and PE) in an isolated system remains constant if there is no air resistance.
02

Find the initial kinetic energy

The initial kinetic energy when the ball is driven can be calculated using the formula: \( KE = \frac{1}{2}mv^2 \), where \( m = 47.0 \, \text{g} = 0.047 \, \text{kg} \) and \( v = 52.0 \, \text{m/s} \). Calculate:\[ KE = \frac{1}{2} \times 0.047 \, \text{kg} \times (52.0 \, \text{m/s})^2 \]\[ KE = 63.7 \, \text{J} \]
03

Calculate the potential energy at maximum height

At the highest point (24.6 m), the potential energy is maximized using the formula \( PE = mgh \), where \( g = 9.81 \, \text{m/s}^2 \) and \( h = 24.6 \, \text{m} \).Calculate:\[ PE = 0.047 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 24.6 \, \text{m} \]\[ PE = 11.34 \, \text{J} \]
04

Use energy conservation to find kinetic energy at highest point

According to the conservation of energy, the initial kinetic energy equals the kinetic energy at the highest point plus the potential energy at that point. Thus, \( KE_{initial} = KE_{highest} + PE_{highest} \).Rearrange to find \( KE_{highest} \):\[ KE_{highest} = KE_{initial} - PE_{highest} \]\[ KE_{highest} = 63.7 \, \text{J} - 11.34 \, \text{J} \]\[ KE_{highest} = 52.36 \, \text{J} \]
05

Use kinetic energy to find speed when 8 m below highest point

First, find the potential energy when the ball is 8 m below the highest point (which is 16.6 m high) using \( PE = mgh \).Calculate:\[ PE = 0.047 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 16.6 \, \text{m} \]\[ PE = 7.67 \, \text{J} \]The kinetic energy at this height will be:\[ KE = KE_{initial} - PE \]\[ KE = 63.7 \, \text{J} - 7.67 \, \text{J} \]\[ KE = 56.03 \, \text{J} \]Now, solve for speed \( v \) using \( KE = \frac{1}{2}mv^2 \):\[ 56.03 \, \text{J} = \frac{1}{2} \times 0.047 \, \text{kg} \times v^2 \]\[ v^2 = \frac{56.03}{0.0235} \, \text{m}^2/\text{s}^2 \]\[ v^2 = 2383.4 \, \text{m}^2/\text{s}^2 \]\[ v = 48.81 \, \text{m/s} \]

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 91影视!

Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. When an object moves, it has the capability to do work by transferring energy to another object through movement. The formula for calculating kinetic energy
  • KE = \( \frac{1}{2}mv^2 \)
  • ''m'' represents mass in kilograms.
  • ''v'' represents velocity in meters per second (m/s).
In the exercise, the golf ball initially possesses kinetic energy as it is driven off the tee at a speed of 52.0 m/s. This initial kinetic energy is calculated by substituting the given mass and velocity into the formula. It tells us about the energy available due to the ball's motion. As the ball rises, it begins to convert some kinetic energy into potential energy.
Potential Energy
Potential energy is the stored energy of position possessed by an object. For objects near the Earth's surface, gravitational potential energy is particularly important. It's given by the formula:
  • PE = \( mgh \)
  • ''m'' is mass in kilograms.
  • ''g'' is the acceleration due to gravity (9.81 m/s虏).
  • ''h'' refers to height above the reference point in meters.
In the exercise, when the golf ball reaches its highest point, it has risen to 24.6 meters above the ground. At this height, the potential energy is maximized since it's stored by its elevated position. Energy conservation tells us that total energy (kinetic + potential) remains the same. Therefore, as the ball gains potential energy by rising, it loses an equivalent amount of kinetic energy. This conversion explains why the ball's speed decreases as it ascends.
Mechanics
Mechanics is a branch of physics focused on the motion of objects and the forces that affect this motion. Within mechanics, we study both the effects of forces on objects and the energy transformations that happen as a result. Energy, whether kinetic or potential, is a crucial aspect of mechanics.
  • Multiple principles of mechanics are often involved in a single problem.
  • Mechanical energy is the sum of kinetic and potential energy.
  • Without external forces like air resistance, mechanical energy remains constant over time.
In the given exercise, the principles of mechanical conservation state that the total sum of kinetic and potential energy of the golf ball does not change (aside from negligible factors like air resistance). This allows us to solve for various unknowns, such as the speed at certain heights during the ball's flight. Mechanics helps us understand and predict the golf ball's behavior in terms of its path and speed at various points.
Projectile Motion
Projectile motion is the movement of an object thrown or projected into the air, subject to only the force of gravity. It鈥檚 a type of parabolic motion described by horizontal and vertical components.
  • Horizontal motion is constant and unaffected by gravity.
  • Vertical motion is uniformly accelerated due to gravity.
  • Projectile motion features changing velocities and varying heights.
In the context of the exercise, the golf ball follows a projectile motion after being hit. This means it has both horizontal motion and vertical motion due to the initial hit and gravity respectively. Its vertical motion, in particular, affects the kinetic and potential energy transformations as it ascends to its highest point and descends. The mechanics of projectile motion give us insights into how the potential and kinetic energy swap places during different phases of its trajectory.

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

The hammer throw is a track-and-field event in which a \(7.3 \mathrm{kg}\) ball (the "hammer"), starting from rest, is whirled around in a circle several times and released. It then moves upward on the familiar curving path of projectile motion. In one throw, the hammer is given a speed of \(29 \mathrm{m} / \mathrm{s} .\) For comparison, a .22 caliber bullet has a mass of \(2.6 \mathrm{g}\) and, starting from rest, exits the barrel of a gun at a speed of \(410 \mathrm{m} / \mathrm{s}\). Determine the work done to launch the motion of (a) the hammer and (b) the bullet.

A Sledding Contest. You are in a sledding contest where you start at a height of \(40.0 \mathrm{m}\) above the bottom of a valley and slide down a hill that makes an angle of \(25.0^{\circ}\) with respect to the horizontal. When you reach the valley, you immediately climb a second hill that makes an angle of \(15.0^{\circ}\) with respect to the horizontal. The winner of the contest will be the contestant who travels the greatest distance up the second hill. You must now choose between using your flat-bottomed plastic sled, or your "Blade Runner," which glides on two steel rails. The hill you will ride down is covered with loose snow. However, the hill you will climb on the other side is a popular sledding hill, and is packed hard and is slick. The two sleds perform very differently on the two surfaces, the plastic one performing better on loose snow, and the Blade Runner doing better on hard-packed snow or ice. The performances of each sled can be quantified in terms of their respective coefficients of kinetic friction on the two surfaces. For the plastic sled: \(\mu=0.17\) on loose snow, and \(\mu=0.15\) on packed snow or ice. For the Blade Runner, \(\mu=0.19\) on loose snow, and \(\mu=0.07\) on packed snow or ice. Assuming the two hills are shaped like inclined planes, and neglecting air resistance, (a) how far does each sled make it up the second hill before stopping? (b) Assuming the total mass of the sled plus rider is \(55.0 \mathrm{kg}\) in both cases, how much work is done by nonconservative forces (over the total trip) in each case?

Under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. When the drive force is shut off, the snowmobile coasts to a halt. The snowmobile and its rider have a mass of 136 kg. Under the influence of a drive force of \(205 \mathrm{N}\), it is moving at a constant velocity whose magnitude is \(5.50 \mathrm{m} / \mathrm{s}\). The drive force is then shut off. Find (a) the distance in which the snowmobile coasts to a halt and (b) the time required to do so.

The drawing shows two frictionless inclines that begin at ground level \((h=0 \mathrm{m})\) and slope upward at the same angle \(\theta .\) One track is longer than the other, however. Identical blocks are projected up each track with the same initial speed \(v_{0}\). On the longer track the block slides upward until it reaches a maximum height \(H\) above the ground. On the shorter track the block slides upward, flies off the end of the track at a height \(H_{1}\) above the ground, and then follows the familiar parabolic trajectory of projectile motion. At the highest point of this trajectory, the block is a height \(H_{2}\) above the end of the track. The initial total mechanical energy of each block is the same and is all kinetic energy. The initial speed of each block is \(v_{0}=7.00 \mathrm{m} / \mathrm{s},\) and each incline slopes upward at an angle of \(\theta=50.0^{\circ} .\) The block on the shorter track leaves the track at a height of \(H_{1}=1.25 \mathrm{m}\) above the ground. Find (a) the height \(H\) for the block on the longer track and (b) the total height \(H_{1}+H_{2}\) for the block on the shorter track.

A 55.0-kg skateboarder starts out with a speed of 1.80 \(\mathrm{m} / \mathrm{s}\). He does \(+80.0 \mathrm{J}\) of work on himself by pushing with his feet against the ground. In addition, friction does -265 J of work on him. In both cases, the forces doing the work are nonconservative. The final speed of the skateboarder is \(6.00 \mathrm{m} / \mathrm{s} .\) (a) Calculate the change \(\left(\Delta \mathrm{PE}=\mathrm{PE}_{\mathrm{f}}-\mathrm{PE}_{0}\right)\) in the gravitational potential energy. (b) How much has the vertical height of the skater changed, and is the skater above or below the starting point?
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Radiation

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Linear Momentum

Read Explanation

Atoms and Radioactivity

Read Explanation
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.