/*! 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 39 A basketball of mass \(0.624 \ma... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 39

A basketball of mass \(0.624 \mathrm{~kg}\) is shot from a vertical height of \(1.2 \mathrm{~m}\) and at a speed of \(20.0 \mathrm{~m} / \mathrm{s}\). After reaching its maximum height, the ball moves into the hoop on its downward path, at \(3.05 \mathrm{~m}\) above the ground. Using the principle of energy conservation, determine how fast the ball is moving just before it enters the hoop.

Short Answer

Expert verified
Answer: The speed of the basketball just before it enters the hoop is approximately 19.1 m/s.

Step by step solution

01

Identifying the known parameters and unknowns

In this problem, we have the following known quantities: the mass of the basketball (m = 0.624 kg), its initial speed (v0 = 20.0 m/s), and the initial height (h0 = 1.2 m). We need to find the final speed of the ball (vf) when it is at a height of 3.05 m above the ground (hf = 3.05 m).
02

Calculate the initial potential and kinetic energy

We first need to calculate the initial kinetic and potential energy of the ball. The kinetic energy (KE) formula is given by: KE = 0.5 * m * v0^2 Note that the gravitational potential energy (PE) formula is given by: PE = m * g * h where g is the acceleration due to gravity (approximately 9.81 m/s虏). Now, calculate the initial kinetic energy (KE_initial) and initial potential energy (PE_initial): KE_initial = 0.5 * 0.624 kg * (20.0 m/s)虏 = 124.8 J PE_initial = 0.624 kg * 9.81 m/s虏 * 1.2 m = 7.33 J
03

Calculate the final potential energy

Next, we need to calculate the final potential energy (PE_final) when the ball is at a height of 3.05 m: PE_final = 0.624 kg * 9.81 m/s虏 * 3.05 m = 18.32 J
04

Apply the principle of energy conservation

Since energy is conserved, we have: initial total energy = final total energy KE_initial + PE_initial = KE_final + PE_final Substituting the known values: 124.8 J + 7.33 J = KE_final + 18.32 J
05

Calculate the final kinetic energy

Now, calculate KE_final: KE_final = (124.8 + 7.33 - 18.32) J = 113.81 J
06

Solve for the final speed of the ball

Recall that the kinetic energy formula is: KE = 0.5 * m * v^2 We already know KE_final. To find the final speed of the ball (vf), we need to rearrange the formula and solve for vf: vf虏 = (2 * KE_final) / m Now, substitute the values: vf虏 = (2 * 113.81 J) / 0.624 kg vf虏 = 364.1 Now, take the square root to find vf: vf = sqrt(364.1) = 19.1 m/s The speed of the basketball just before it enters the hoop is approximately 19.1 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
Understanding kinetic energy is crucial in solving physics problems involving moving objects. Kinetic energy (\textbf{KE}), often visualized as the energy of motion, is quantified by the equation \[ KE = \frac{1}{2} m v^2 \]
In this formula, \(m\) represents the mass of the object and \(v\) is its velocity. An object gains kinetic energy as it speeds up and loses it when slowing down. For instance, in the basketball problem, we initially gathered the kinetic energy using its mass and initial speed, depicting how much energy is tied up in its movement.
Potential Energy
Potential energy (\textbf{PE}) relates to the position or condition of an object. Gravitational potential energy is the most relevant form for objects near Earth's surface, calculated through the equation \[ PE = m g h \]
Where \(g\) is the acceleration due to gravity (\(9.81 \text{m/s}^2\)), and \(h\) is the object's height above a reference point. The basketball's potential energy at different heights showcases this object's stored energy due to gravity, convertible to kinetic energy when the basketball falls.
Conservation of Energy
The principle of conservation of energy states that within a closed system, energy can neither be created nor destroyed, only transformed from one form to another. This key principle allows us to set up an equation in our basketball problem that equates the sum of kinetic and potential energy at one point with that sum at another point. Thus, by knowing the initial total energy, we can deduce the final kinetic energy as the basketball falls toward the hoop. The formula we used, \[ KE_{\text{initial}} + PE_{\text{initial}} = KE_{\text{final}} + PE_{\text{final}} \],
effectively captures the essence of energy conservation in the context of the problem.
Kinematics in Physics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces causing such motion. It encompasses concepts like velocity, acceleration, and the physics of projectiles. In addressing our basketball problem, kinematics intertwines with energy principles. For example, the ball鈥檚 height at any point during its journey affects both its potential energy and its kinetic energy, demonstrating a kinematic relationship between position and velocity. Solving for the basketball's final speed required us to piece this relationship together鈥攁nd in doing so, we practiced the applications of kinematics in the realm of energy conservation.

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

A 0.100 -kg ball is dropped from a height of \(1.00 \mathrm{~m}\) and lands on a light (approximately massless) cup mounted on top of a light, vertical spring initially at its equilibrium position. The maximum compression of the spring is to be \(10.0 \mathrm{~cm}\). a) What is the required spring constant of the spring? b) Suppose you ignore the change in the gravitational energy of the ball during the 10 -cm compression. What is the percentage difference between the calculated spring constant for this case and the answer obtained in part (a)?

A 5.00 -kg ball of clay is thrown downward from a height of \(3.00 \mathrm{~m}\) with a speed of \(5.00 \mathrm{~m} / \mathrm{s}\) onto a spring with \(k=\) \(1600 . \mathrm{N} / \mathrm{m} .\) The clay compresses the spring a certain maximum amount before momentarily stopping a) Find the maximum compression of the spring. b) Find the total work done on the clay during the spring's compression.

An arrow is placed on a bow, the bowstring is pulled back, and the arrow is shot straight up into the air; the arrow then comes back down and sticks into the ground. Describe all of the changes in work and energy that occur.

A 1.00 -kg block is resting against a light, compressed spring at the bottom of a rough plane inclined at an angle of \(30.0^{\circ}\); the coefficient of kinetic friction between block and plane is \(\mu_{\mathrm{k}}=0.100 .\) Suppose the spring is compressed \(10.0 \mathrm{~cm}\) from its equilibrium length. The spring is then released, and the block separates from the spring and slides up the incline a distance of only \(2.00 \mathrm{~cm}\) beyond the spring's normal length before it stops. Determine a) the change in total mechanical energy of the system and b) the spring constant \(k\).

A 1.00 -kg block compresses a spring for which \(k=\) 100. \(\mathrm{N} / \mathrm{m}\) by \(20.0 \mathrm{~cm}\) and is then released to move across a horizontal, frictionless table, where it hits and compresses another spring, for which \(k=50.0 \mathrm{~N} / \mathrm{m}\). Determine a) the total mechanical energy of the system, b) the speed of the mass while moving freely between springs, and c) the maximum compression of the second spring.
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Rotational Dynamics

Read Explanation

Atoms and Radioactivity

Read Explanation

Thermodynamics

Read Explanation

Wave Optics

Read Explanation

Electricity and Magnetism

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.