/*! 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 17 The hammer throw is a track-and-... [FREE SOLUTION] | 91Ó°ÊÓ

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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 with 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: 3070.35 J, (b) Bullet: 218.53 J.

Step by step solution

01

Understand Work-Energy Principle

The work done on an object is given by its change in kinetic energy. The formula to calculate work done based on the initial and final speeds is \( W = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the speed.
02

Calculate Work Done on the Hammer

The hammer has a mass \( m_1 = 7.3 \) kg and is given a speed \( v_1 = 29 \) m/s. Use the work-energy principle formula: \[ W_1 = \frac{1}{2} \times 7.3 \times (29)^2 \] Calculate to find \( W_1 \).
03

Solve for the Work on the Hammer

Substitute the values into the formula:\[ W_1 = \frac{1}{2} \times 7.3 \times 841 \] \[ W_1 = 3070.35 \] J. The work done on the hammer is 3070.35 Joules.
04

Convert Bullet Mass to Kilograms

The bullet's mass is given in grams, so convert it to kilograms: \( 2.6 \) g = \( 0.0026 \) kg.
05

Calculate Work Done on the Bullet

The bullet has a mass \( m_2 = 0.0026 \) kg and is given a speed \( v_2 = 410 \) m/s. Use the work-energy principle formula: \[ W_2 = \frac{1}{2} \times 0.0026 \times (410)^2 \] Calculate to find \( W_2 \).
06

Solve for the Work on the Bullet

Substitute the values into the formula:\[ W_2 = \frac{1}{2} \times 0.0026 \times 168100 \] \[ W_2 = 218.53 \] J. The work done on the bullet is 218.53 Joules.

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Key Concepts

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

Understanding Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. Whenever an object is moving, it has kinetic energy, which depends on two main factors: the mass of the object and its velocity. The formula to calculate kinetic energy is:
  • \( KE = \frac{1}{2}mv^2 \)
where \( m \) is the mass of the object and \( v \) is its velocity. This means that the heavier an object is and the faster it is moving, the more kinetic energy it has.
For example, in the hammer throw event, the hammer is initially stationary. As the athlete spins and releases it, the hammer gains velocity and thus kinetic energy. Similarly, a bullet gains kinetic energy when fired. Understanding this energy transformation helps us determine how much work is done to set these objects in motion.
Work, in physics, is the process of energy transfer. The work done on an object is equal to the change in kinetic energy. If you want to calculate how much work was done to get an object moving, you can use the work-energy principle. This principle states:
  • Work done \( W = \text{Change in Kinetic Energy} = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 \)
where \( u \) is the initial velocity (often zero if the object starts from rest), and \( v \) is the final velocity.
Projectile Motion Basics
Projectile motion refers to the motion of an object that is thrown or projected into the air at an angle. This type of motion is characterized by both horizontal and vertical components. The dynamics of projectile motion are influenced by factors such as:
  • The initial velocity of the projectile
  • The angle of launch
  • The acceleration due to gravity
In a typical hammer throw situation, once the hammer is released, it follows a curved path known as a parabola. This trajectory results from the combination of two types of motion at play: a constant velocity in the horizontal direction and an accelerated motion in the vertical direction due to gravity.
The key points about projectile motion include:
  • The horizontal motion is uniform, which means it maintains a constant speed since air resistance is often negligible.
  • Vertical motion is non-uniform as it is influenced by gravity, causing the object to accelerate downwards.
  • The highest point in the trajectory is known as the apex, where the vertical velocity component is zero.
Understanding projectile motion is crucial for analyzing the hammer's path once it is released. It helps predict where the hammer will land given its initial speed and launch angle.
Importance of Conversion of Units
Physics involves various quantities that are often given in different units. It is vital to convert these units to ensure calculations are accurate. When dealing with mass and velocity, typical units are kilograms and meters per second respectively.
In the case study provided, one must first convert the mass of the bullet from grams to kilograms before applying the work-energy principle. The conversion factor is:
  • 1 gram = 0.001 kilograms
This conversion is necessary because the standard unit of mass in the International System of Units (SI) is the kilogram. For instance, when calculating the work done on the bullet, converting 2.6 grams to 0.0026 kilograms ensures consistency across calculations. Similarly, ensuring that all velocity measurements are in meters per second helps maintain the integrity of the calculations using kinetic energy formulas.
Using consistent units allows for accurate applications of physics formulas and principles. Different contexts may require unit conversions, so it’s essential to be mindful of these conversions to avoid errors.

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Most popular questions from this chapter

A sled is being pulled across a horizontal patch of snow. Friction is negligible. The pulling force points in the same direction as the sled's displacement, which is along the \(+x\) axis. As a result, the kinetic energy of the sled increases by \(38 \% .\) By what percentage would the sled's kinetic energy have increased if this force had pointed \(62^{\circ}\) above the \(+x\) axis?

A truck is traveling at \(11.1 \mathrm{~m} / \mathrm{s}\) down a hill when the brakes on all four wheels lock. The hill makes an angle of \(15.0^{\circ}\) with respect to the horizontal. The coefficient of kinetic friction between the tires and the road is \(0.750\). How far does the truck skid before coming to a stop?

A \(1200-\mathrm{kg}\) car is being driven up a \(5.0^{\circ}\) hill. The frictional force is directed opposite to the motion of the car and has a magnitude of \(f=524 \mathrm{~N}\). A force \(\overrightarrow{\mathrm{F}}\) is applied to the car by the road and propels the car forward. In addition to these two forces, two other forces act on the car: its weight \(\overrightarrow{\mathrm{W}}\) and the normal force \(\overrightarrow{\mathrm{F}}_{\mathrm{N}}\) directed perpendicular to the road surface. The length of the road up the hill is \(290 \mathrm{~m}\). What should be the magnitude of \(\overrightarrow{\mathbf{F}}\), so that the net work done by all the forces acting on the car is \(+150 \mathrm{~kJ}\) ?

A \(2.00\) -kg rock is released from rest at a height of \(20.0 \mathrm{~m}\). Ignore air resistance and determine the kinetic energy, gravitational potential energy, and total mechanical energy at each of the following heights: \(20.0,10.0\), and \(0 \mathrm{~m}\).

When an \(81.0\) - \(\mathrm{kg}\) adult uses a spiral staircase to climb to the second floor of his house, his gravitational potential energy increases by \(2.0 \times 10^{3} \mathrm{~J}\). By how much does the potential energy of an \(18.0\) - \(\mathrm{kg}\) child increase when the child climbs a normal staircase to the second floor?

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