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Chapter 6: Problem 19

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.

Short Answer

Expert verified
(a) Hammer: 3069.65 J, (b) Bullet: 218.53 J.

Step by step solution

01

Understand Work-Energy Principle

The work done on an object is equal to the change in its kinetic energy. The formula for work is given by: \( W = \Delta KE = \frac{1}{2} mv^2 - \frac{1}{2} mu^2 \), where \( m \) is the mass of the object, \( v \) is the final velocity, and \( u \) is the initial velocity. For both the hammer and the bullet, they start from rest, so \( u = 0 \).
02

Calculate Work Done on the Hammer

Substituting the given values into the work formula for the hammer: \( m = 7.3 \text{ kg} \) and \( v = 29 \text{ m/s} \). Therefore, the work done is calculated as follows: \( W = \frac{1}{2} \times 7.3 \times 29^2 = \frac{1}{2} \times 7.3 \times 841 \).
03

Solve for Work Done on the Hammer

Calculate the numerical value of the work done on the hammer. \( W = 0.5 \times 7.3 \times 841 = 3069.65 \text{ J} \). This is the work done to launch the hammer.
04

Calculate Work Done on the Bullet

Substituting the given values into the work formula for the bullet: Convert the mass from grams to kilograms by dividing by 1000, so \( m = 2.6 \text{ g} = 0.0026 \text{ kg} \) and \( v = 410 \text{ m/s} \). Therefore, the work done is calculated as follows: \( W = \frac{1}{2} \times 0.0026 \times 410^2 = \frac{1}{2} \times 0.0026 \times 168100 \).
05

Solve for Work Done on the Bullet

Calculate the numerical value of the work done on the bullet. \( W = 0.5 \times 0.0026 \times 168100 = 218.53 \text{ J} \). This is the work done to launch the bullet.

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Key Concepts

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

Projectile Motion
Projectile motion is a fascinating aspect of physics that occurs in objects that are thrown or launched into the air and subject to the force of gravity. These objects, such as the hammer in a hammer throw, follow a curved path, known as a trajectory. This trajectory is parabolic in nature due to the influences of both horizontal and vertical motion acting simultaneously.
  • Horizontal Motion: This part of the motion occurs at a constant velocity. It is unaffected by gravity because the act of gravity works perpendicular to the horizontal direction.
  • Vertical Motion: Here, the object experiences acceleration due to gravity, leading to variations in velocity. This results in the rise, peak, and eventual fall of the object.
Understanding projectile motion is crucial as it allows us to predict the path an object will take after being launched. For instance, in sports like hammer throw, athletes can adjust their technique to achieve the optimum trajectory for maximum distance.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It is an integral part of the work-energy principle, where work done on an object results in a change in its kinetic energy. The formula for kinetic energy is given by:\[ KE = \frac{1}{2} m v^2 \]
  • Mass \(m\): The mass of the object being moved plays a significant role in determining how much kinetic energy it possesses.
  • Velocity \(v\): Even more critical is the object's velocity. Since velocity is squared in the formula, a small increase in speed results in a much larger increase in kinetic energy.
By applying this formula, we can determine how much work is done in accelerating an object from rest to a certain speed. In our exercise, for both the hammer and bullet, this principle helps calculate the work exerted to achieve their final velocities.
Mass Conversion
In physics, understanding mass conversion is vital when dealing with various calculations, especially when the units need consistency. Typically, we encounter mass in grams, kilograms, or other units, necessitating conversions to keep calculations accurate. The standard approach is converting grams to kilograms, as the metric system base unit for mass in scientific work is kilograms.
  • A straightforward conversion formula is utilized: \( 1 \text{ kg} = 1000 \text{ g} \).
  • When dealing with problems like the bullet鈥檚 mass in the exercise, converting from 2.6 grams to kilograms becomes essential to seamlessly apply the kinetic energy formula \( W = \frac{1}{2} mv^2 \).
By ensuring mass units are converted and consistent, calculations become simpler, and errors are minimized. This principle is especially useful in exercises involving energy, where precision is crucial for accurately determining outcomes.

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Most popular questions from this chapter

A pendulum consists of a small object hanging from the ceiling at the end of a string of negligible mass. The string has a length of \(0.75 \mathrm{m}\). With the string hanging vertically, the object is given an initial velocity of \(2.0 \mathrm{m} / \mathrm{s}\) parallel to the ground and swings upward on a circular arc. Eventually, the object comes to a momentary halt at a point where the string makes an angle \(\theta\) with its initial vertical orientation and then swings back downward. Find the angle \(\theta\)

You are working out on a rowing machine. Each time youpull the rowing bar (which simulates the oars) toward you, it moves a distance of \(1.2 \mathrm{m}\) in a time of \(1.5 \mathrm{s}\). The readout on the display indicates that the average power you are producing is 82 W. What is the magnitude of the force that you exert on the handle?

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?

The (nonconservative) force propelling a \(1.50 \times 10^{3}-\mathrm{kg}\) car up a mountain road does \(4.70 \times 10^{6} \mathrm{J}\) of work on the car. The car starts from rest at sea level and has a speed of \(27.0 \mathrm{m} / \mathrm{s}\) at an altitude of \(2.00 \times 10^{2} \mathrm{m}\) above sea level. Obtain the work done on the car by the combined forces of friction and air resistance, both of which are nonconservative forces.

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?
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