/*! 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 14 A golf club strikes a 0.045-kg g... [FREE SOLUTION] | 91Ó°ÊÓ

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A golf club strikes a 0.045-kg golf ball in order to launch it from the tee. For simplicity, assume that the average net force applied to the ball acts parallel to the ball's motion, has a magnitude of \(6800 \mathrm{N},\) and is in contact with the ball for a distance of \(0.010 \mathrm{m}\). With what speed does the ball leave the club?

Short Answer

Expert verified
The ball leaves the club with a speed of approximately 54.98 m/s.

Step by step solution

01

Understanding the problem

The problem asks us to find the speed of a golf ball given a force and a contact distance. We are given the mass of the ball, the average force applied, and the distance over which this force is exerted.
02

Use the Work-Energy Principle

According to the work-energy principle, the work done on an object is equal to the change in its kinetic energy. The work done, \( W \), is calculated using the formula \( W = F \cdot d \), where \( F \) is the force and \( d \) is the distance over which the force is applied.
03

Calculate the Work Done

Calculate the work done on the ball: \( W = 6800 \, \text{N} \times 0.010 \, \text{m} = 68 \, \text{J} \). This is the amount of energy transferred to the ball.
04

Relate Work to Kinetic Energy

The work done on the ball increases its kinetic energy. The formula for kinetic energy \( KE \) is \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity. Equate the work done to the kinetic energy: \( 68 \, \text{J} = \frac{1}{2} \times 0.045 \, \text{kg} \times v^2 \).
05

Solve for Ball's Velocity

Rearrange the kinetic energy equation to solve for \( v \). \[ 68 = \frac{1}{2} \times 0.045 \times v^2 \]\[ v^2 = \frac{68}{0.0225} \]\[ v^2 = 3022.2222 \]\[ v = \sqrt{3022.2222} \]\[ v \approx 54.98 \, \text{m/s} \]. Thus, the speed of the ball when it leaves the club is approximately 54.98 m/s.

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Key Concepts

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

Kinetic Energy
Kinetic energy is a form of energy that an object possesses due to its motion. It is directly related to the speed and mass of the object. The more massive an object and the faster it moves, the greater its kinetic energy. The formula for kinetic energy is given by \[KE = \frac{1}{2}mv^2\]where
  • \( m \) is the mass of the object
  • \( v \) is the velocity of the object
When considering the kinetic energy of an object, it's essentially the work needed to accelerate the object from rest to a given speed.
A key idea here is that kinetic energy is proportional to the square of the velocity. This means that if you double the velocity of an object, its kinetic energy increases by a factor of four.
This relationship is helpful in understanding why even small increases in speed can significantly increase kinetic energy. In this exercise, the work done by the golf club on the ball translates into an increase in the ball's kinetic energy. As a result, we can calculate the ball's speed just by knowing the work exerted.
Force and Motion
Force plays a crucial role in physics to cause or change motion. It acts as a push or a pull on an object. Forces can either make an object start moving, speed up, slow down, or change direction.
According to Newton's Second Law, the relationship between force, mass, and acceleration is \[F = ma\] where
  • \( F \) is the force applied
  • \( m \) is the mass of the object
  • \( a \) is the acceleration produced
The direction of the force directly influences the direction of the motion. In our problem, the force is applied in the same direction as the motion of the golf ball.
This means that the force contributes entirely to moving the ball forward.
Additionally, since the force acts over a distance, we also consider the work done by the force. The concept of work here is critical, as it links force with energy. Work, defined as force times distance, converts the force along the distance into kinetic energy, propelling the golf ball.
Problem-Solving in Physics
Problem-solving in physics often involves systematically breaking down a problem into manageable parts.
Let's outline a general approach:
  • Understand the Problem: Identify what is given and what needs to be found. This foundational step sets the stage for applying relevant principles.
  • Apply Physics Principles: Use laws and equations like the work-energy principle, Newton's laws, etc., relevant to the problem. Selecting the right principle is crucial.
  • Perform Calculations: Substitute known values into the equations and perform the necessary mathematical operations. This step requires careful attention to units and proper arithmetic.
  • Verify Results: Check if your answers make sense logically and physically. This includes considering if the units make sense and if the quantity's final value is reasonable.
This structured approach helps in acquiring clear solutions and ensures that no critical details are overlooked. In more technical problems like calculating the speed of the golf ball, following these steps will guide one through the problem-solving process effectively and efficiently. The work-energy principle applied here is a cornerstone for linking mechanical principles to solve such dynamic problems.

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Most popular questions from this chapter

A gymnast is swinging on a high bar. The distance between his waist and the bar is \(1.1 \mathrm{m},\) as the drawing shows. At the top of the swing his speed is momentarily zero. Ignoring friction and treating the gymnast as if all of his mass is located at his waist, find his speed at the bottom of the swing.

Bicyclists in the Tour de France do enormous amounts of work during a race. For example, the average power per kilogram generated by seven-time-winner Lance Armstrong \((m=75.0 \mathrm{kg})\) is \(6.50 \mathrm{W}\) per kilogram of his body mass. (a) How much work does he do during a 135 -km race in which his average speed is \(12.0 \mathrm{m} / \mathrm{s} ?\) (b) Often, the work done is expressed in nutritional Calories rather than in joules. Express the work done in part (a) in terms of nutritional Calories, noting that 1 joule \(=2.389 \times 10^{-4}\) nutritional Calories.

A 35-kg girl is bouncing on a trampoline. During a certain interval after she leaves the surface of the trampoline, her kinetic energy decreases to 210 J from 440 J. How high does she rise during this interval? Neglect air resistance.

A \(16-\mathrm{kg}\) sled is being pulled along the horizontal snow-covered ground by a horizontal force of \(24 \mathrm{N}\). Starting from rest, the sled attains a speed of \(2.0 \mathrm{m} / \mathrm{s}\) in \(8.0 \mathrm{m} .\) Find the coefficient of kinetic friction between the runners of the sled and the snow.

A basketball player makes a jump shot. The \(0.600-\mathrm{kg}\) ball is released at a height of \(2.00 \mathrm{m}\) above the floor with a speed of \(7.20 \mathrm{m} / \mathrm{s}\). The ball goes through the net \(3.10 \mathrm{m}\) above the floor at a speed of \(4.20 \mathrm{m} / \mathrm{s}\). What is the work done on the ball by air resistance, a nonconservative force?

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