/*! 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-\m... [FREE SOLUTION] | 91Ó°ÊÓ

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A golf club strikes a \(0.045-\mathrm{kg}\) 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 golf ball leaves the club at about 54.8 m/s.

Step by step solution

01

Understanding the Problem

We need to find the speed of the golf ball as it leaves the club. The problem describes the force applied to the ball, the distance over which the force acts, and the mass of the ball.
02

Applying the Work-Energy Principle

The work-energy principle states that the work done on an object is equal to the change in kinetic energy of the object. The work done by the force is given by the formula:\[ W = F \cdot d \]where \( F = 6800 \, \mathrm{N} \) is the force and \( d = 0.010 \, \mathrm{m} \) is the distance.
03

Calculating the Work Done

Substitute the values into the work formula:\[ W = 6800 \, \mathrm{N} \times 0.010 \, \mathrm{m} = 68 \, \mathrm{J} \]This work is responsible for the change in kinetic energy of the ball.
04

Relating Work to Kinetic Energy

The work done on the ball (68 J) equals its kinetic energy change because it starts from rest:\[ 68 \, \mathrm{J} = \frac{1}{2}mv^2 \]where \( m = 0.045 \, \mathrm{kg} \) and \( v \) is the speed.
05

Solving for the Final Velocity

Rearrange the kinetic energy equation to solve for \( v \):\[ v = \sqrt{\frac{2W}{m}} \]Substitute the known values:\[ v = \sqrt{\frac{2 \times 68 \, \mathrm{J}}{0.045 \, \mathrm{kg}}} \]Calculate the result:\[ v \approx 54.8 \, \mathrm{m/s} \]
06

Solution Verification

Double-check the calculations and units to ensure the procedure is correct and consistent with the problem conditions.

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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 fundamental concept in physics that describes the energy an object possesses due to its motion. This energy depends on two key factors:
  • the mass of the object
  • the velocity of the object
The mathematical formula for kinetic energy is given by:
\[ KE = \frac{1}{2}mv^2 \]Here, \(m\) represents the mass and \(v\) stands for velocity.
Kinetic energy changes when the speed or mass of an object changes. In the case of our golf ball problem, once the golf club strikes the ball, the force acts over a distance, doing work on the ball, which then changes its kinetic energy. Very importantly, the ball starts from rest, so initially its kinetic energy is zero. As it gains speed, its kinetic energy increases, which means work has been effectively done on it.
Net Force
Net force refers to the overall force acting on an object when all individual forces acting on it are considered. When calculating net force, it's important to account for both magnitude and direction.
  • The net force is responsible for changes in the motion of an object.
  • In our particular problem, the force applied by the golf club is 6800 N and acts parallel to the ball's motion.
  • This is a straightforward scenario where only one significant force is acting, simplifying our calculations greatly.
Understanding net force improves our comprehension of how external influences cause changes in velocity and kinetic energy, further linking back to the work-energy principle.
Physics Problem Solving
Solving physics problems involves a structured approach that helps break down complex situations into manageable parts. In our exercise, a logical progression was used:
  • Start by understanding what is asked and identifying known values and unknowns.
  • Apply relevant physical principles, like the work-energy principle, to relate known measurements to what we need to find.
  • Step-by-step calculations, such as finding the work done on the ball, are executed precisely.
  • Rechecking calculations ensures accuracy and that no steps were skipped.
Building this systematic habit allows you to tackle a variety of physics problems with confidence and precision.
Golf Ball Dynamics
Golf ball dynamics involves understanding both the forces at play and the resultant motion. When a golf club hits a ball:
  • The point of contact and the force applied determine the ball's initial trajectory and speed.
  • In our exercise, the force is applied over a small distance of 0.010 meters, yet it results in significant kinetic energy due to the magnitude of the force.
  • After being struck, the ball's speed of 54.8 m/s tells us how effectively the golf club transferred energy to the ball during impact.
These principles can also help enhance techniques in sports, highlighting the crossover between physics and practical applications in real-life activities.

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Most popular questions from this chapter

A car accelerates uniformly from rest to \(20.0 \mathrm{m} / \mathrm{s}\) in \(5.6 \mathrm{s}\) along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car if (a) the weight of the car is \(9.0 \times 10^{3} \mathrm{N}\) and (b) the weight of the car is \(1.4 \times 10^{4} \mathrm{N}\).

A 63-kg skier coasts up a snow-covered hill that makes an angle of \(25^{\circ}\) with the horizontal. The initial speed of the skier is \(6.6 \mathrm{m} / \mathrm{s}\). After coasting \(1.9 \mathrm{m}\) up the slope, the skier has a speed of \(4.4 \mathrm{m} / \mathrm{s}\). (a) Find the work done by the kinetic frictional force that acts on the skis. (b) What is the magnitude of the kinetic frictional force?

A \(1.00 \times 10^{2}-\mathrm{kg}\) crate is being pushed across a horizontal floor by a force \(\overrightarrow{\mathbf{P}}\) that makes an angle of \(30.0^{\circ}\) below the horizontal. The coefficient of kinetic friction is \(0.200 .\) What should be the magnitude of \(\overrightarrow{\mathbf{P}},\) so that the net work done by it and the kinetic frictional force is zero?

A 67.0 -kg person jumps from rest off a \(3.00-\mathrm{m}\) -high tower straight down into the water. Neglect air resistance. She comes to rest \(1.10 \mathrm{m}\) under the surface of the water. Determine the magnitude of the average force that the water exerts on the diver. This force is nonconservative.

A husband and wife take turns pulling their child in a wagon along a horizontal sidewalk. Each exerts a constant force and pulls the wagon through the same displacement. They do the same amount of work, but the husband's pulling force is directed \(58^{\circ}\) above the horizontal, and the wife's pulling force is directed \(38^{\circ}\) above the horizontal. The husband pulls with a force whose magnitude is \(67 \mathrm{N}\). What is the magnitude of the pulling force exerted by his wife?

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