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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

Two pole-vaulters just clear the bar at the same height. The first lands at a speed of \(8.90 \mathrm{m} / \mathrm{s}\), and the second lands at a speed of \(9.00 \mathrm{m} / \mathrm{s}\). The first vaulter clears the bar at a speed of \(1.00 \mathrm{m} / \mathrm{s}\). Ignore air resistance and friction and determine the speed at which the second vaulter clears the bar.

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

A slingshot fires a pebble from the top of a building at a speed of \(14.0 \mathrm{m} / \mathrm{s} .\) The building is \(31.0 \mathrm{m}\) tall. Ignoring air resistance, find the speed with which the pebble strikes the ground when the pebble is fired (a) horizontally, (b) vertically straight up, and (c) vertically straight down.

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?

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.
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