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Chapter 6: Problem 14

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鈥檚 motion, has a magnitude of 6800 N, and is in contact with the ball for a distance of 0.010 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

Identify Given Information

Identify the values given in the problem statement. The mass of the ball is \( m = 0.045 \; \text{kg} \), the force applied is \( F = 6800 \; \text{N} \), and the distance over which the force is applied is \( d = 0.010 \; \text{m} \).
02

Understand the Concept of Work-Energy Principle

The work done on the ball by the force is transformed into kinetic energy. We can link this to find the speed of the ball as it leaves the club using the work-energy principle: \( W = \Delta KE \).
03

Calculate Work Done

Calculate the work done on the ball using the formula \( W = F \cdot d \). Substitute the given values: \( W = 6800 \; \text{N} \times 0.010 \; \text{m} = 68 \; \text{J} \).
04

Set Up Kinetic Energy Equation

The change in kinetic energy is equal to the work done on the ball. Set up the equation for kinetic energy: \( \Delta KE = \frac{1}{2} m v^2 - 0 \). Since the ball started from rest, initial kinetic energy is zero, so \( \Delta KE = \frac{1}{2} m v^2 \).
05

Solve for Velocity

Substitute \( \Delta KE = W = 68 \; \text{J} \) into the kinetic energy equation: \( 68 = \frac{1}{2} \times 0.045 \times v^2 \). Solve for \( v \): \( v^2 = \frac{136}{0.045} = 3022.22 \), and \( v = \sqrt{3022.22} \approx 54.98 \; \text{m/s} \).

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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 fundamental concept in physics that describes the energy possessed by an object due to its motion. It is given by the formula:
  • \( KE = \frac{1}{2} m v^2 \)
where \( m \) is the mass of the object, and \( v \) is its velocity. When an object is at rest, its kinetic energy is zero, but as it starts moving, the energy increases.
In the context of a golf ball being hit by a club, the impact on the ball provides the energy needed for the ball to start moving. This energy is transferred from the club to the ball, increasing the golf ball's kinetic energy. The calculation of kinetic energy helps us determine how fast the ball is moving after being hit.
Average Net Force
Average net force is an essential part of understanding how objects are set in motion. It describes the total force acting on an object averaged over the time of contact or the distance over which it acts.
In the case of a golf club striking a ball, we consider the average net force to be the constant force exerted on the ball. This force is what causes the change in the ball鈥檚 velocity.
The magnitude of the force applied, combined with the distance over which it acts, contributes directly to the work done on the ball. This work is then converted entirely into the ball's kinetic energy, allowing it to leave the club with a certain speed.
Velocity Calculation
Velocity calculation involves determining the speed and direction of an object. In scenarios like striking a golf ball, we calculate the velocity to understand how quickly and in which direction the ball will travel after impact.
To find the velocity after the club hits the ball, we can use the work-energy principle, which states that the work done on an object is equal to its change in kinetic energy. Using:
  • \( W = \Delta KE \)
  • \( W = \frac{1}{2} m v^2 \)
from there, solving for \( v \), we substitute the work (energy transferred from the net force) into the kinetic energy formula. Calculating this is how we arrive at the ball's final velocity, giving us the necessary insight into how force and energy translate into movement.
Golf Ball Physics
Golf ball physics encompasses understanding how physics principles apply to the motion and behavior of golf balls. By striking the ball with a club, we delve into concepts like force, energy, and motion.
The crucial elements include:
  • The mass of the ball - affects how much force is needed to achieve a certain speed.
  • The distance and direction the ball travels - influenced by its initial velocity imparted by the club.
  • The resistance the ball encounters (like air resistance) - which can also alter its path and speed.
Grasping how these factors play together allows not only for the calculation of velocity but also for predictions about the ball's trajectory and behavior once it leaves the club.

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Most popular questions from this chapter

The surfer in the photo is catching a wave. Suppose she starts at the top of the wave with a speed of 1.4 m/s and moves down the wave until her speed increases to 9.5 m/s. The drop in her vertical height is 2.7 m. If her mass is 59 kg, how much work is done by the (non conservative) force of the wave?

The drawing shows two friction less inclines that begin at ground level \((h=0 m)\) 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 \(v_{0}=7.00 m /s,\) 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 \(H_{1}=1.25 m\) 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.

The hammer throw is a track-and-field event in which a 7.3-kg ball (the 鈥渉ammer鈥�), 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 m/s. For comparison, a .22 caliber bullet has a mass of 2.6 g and, starting from rest, exits the barrel of a gun at a speed of 410 m/s. Determine the work done to launch the motion of (a) the hammer and (b) the bullet.

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 0.60-kg basketball is dropped out of a window that is 6.1 m above the ground. The ball is caught by a person whose hands are 1.5 m above the ground. (a) How much work is done on the ball by its weight? What is the gravitational potential energy of the basketball, relative to the ground, when it is (b) released and (c) caught? (d) How is the change \(\left(PE_{\mathrm{f}}-PE_{0}\right)\) in the ball's gravitational potential energy related to the work done by its weight?
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