/*! 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 49 Typically, a bullet leaves a sta... [FREE SOLUTION] | 91Ó°ÊÓ

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Typically, a bullet leaves a standard 45 -caliber pistol (5.0-in. barrel) at a speed of \(262 \mathrm{~m} / \mathrm{s}\). If it takes \(1 \mathrm{~ms}\) to traverse the barrel, determine the average acceleration experienced by the \(16.2-\mathrm{g}\) bullet within the gun, and then compute the average force exerted on it.

Short Answer

Expert verified
The average acceleration is \(262,000 \mathrm{~m/s^2}\), and the average force on the bullet is \(4,244.4 \mathrm{~N}\).

Step by step solution

01

Understanding Given Values

We are given a bullet speed of \(262 \mathrm{~m/s}\), a time of \(1 \mathrm{~ms} = 0.001 \mathrm{~s}\), and a bullet mass of \(16.2 \mathrm{~g} = 0.0162 \mathrm{~kg}\). We need to find the average acceleration and force on the bullet.
02

Calculate the Average Acceleration

The formula for acceleration is \(a = \frac{\Delta v}{t}\), where \(\Delta v\) is the change in velocity and \(t\) is the time taken. Here, \(\Delta v = 262 \mathrm{~m/s}\) and \(t = 0.001 \mathrm{~s}\). Therefore, \(a = \frac{262 \mathrm{~m/s}}{0.001 \mathrm{~s}} = 262,000 \mathrm{~m/s^2}\).
03

Calculate the Average Force

Using Newton's second law, \(F = ma\), where \(F\) is the force, \(m\) is the mass (\(0.0162 \mathrm{~kg}\)), and \(a\) is the acceleration (\(262,000 \mathrm{~m/s^2}\)). Therefore, \(F = 0.0162 \mathrm{~kg} \times 262,000 \mathrm{~m/s^2} = 4,244.4 \mathrm{~N}\).

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Key Concepts

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

Average Acceleration
To determine the average acceleration experienced by an object, we must understand the concept of change in velocity over a specific period of time. Average acceleration is defined as the rate at which velocity changes. It's denoted by the formula:
  • \[ a = \frac{\Delta v}{t} \]
  • Where \(\Delta v\) is the change in velocity and \(t\) is the time taken.
In the context of our bullet problem, the bullet's velocity changes from zero (since we assume it starts from rest) to \(262 \, \mathrm{m/s}\) as it travels through a \(5.0\)-inch barrel in \(1 \, \mathrm{ms}\) (or \(0.001 \, \mathrm{s}\)). The substitution of these values gives us an average acceleration of \(262,000 \, \mathrm{m/s^2}\). This high value is typical for such rapid change in speed over a short time span, as experienced in firearms.
Newton's Second Law
Newton's Second Law is a core principle in physics, describing how the velocity of an object changes when it is subjected to an external force. This law is succinctly expressed by the equation:
  • \[ F = ma \]
  • where \(F\) is the force applied, \(m\) is the mass of the object, and \(a\) is the acceleration.
This is especially pertinent in calculating the force on a bullet as it is accelerated within the barrel. When we know the mass of the bullet (\(0.0162 \, \mathrm{kg}\)) and its acceleration (\(262,000 \, \mathrm{m/s^2}\)), we can compute the force using Newton's second law. This results in a force of \(4,244.4 \, \mathrm{N}\). This immense force is what propels the bullet at such a high speed out of the gun.
Kinematics
Kinematics is a branch of physics that deals with the motion of objects without considering the forces that cause the motion. In the context of our problem, we focus on the bullet's movement within the barrel, described by its initial and final velocities, the time it takes to traverse the barrel, and its acceleration. The key parameters include:
  • Initial Velocity: Often \(0 \, \mathrm{m/s}\) when starting from rest
  • Final Velocity: \(262 \, \mathrm{m/s}\) for our bullet
  • Time Span: \(0.001 \, \mathrm{s}\), the time taken to cover the barrel length
  • Acceleration: derived from change in velocity \(\Delta v\) over time \(t\)
These aspects define the bullet's motion as it speeds up from the back to the front of the gun barrel, allowing for the calculation of acceleration and force.
Force Calculation
Force calculation, as derived from the principles of Newton's Second Law, is crucial for understanding how fast and powerfully a bullet exits a firearm. Using the formula:
  • \[ F = ma \]
We first determine the bullet's acceleration using kinematics principles, and then apply it to calculate the force. The mass of the bullet is \(0.0162 \, \mathrm{kg}\), and the acceleration is \(262,000 \, \mathrm{m/s^2}\) as computed previously. Thus, the average force exerted on the bullet inside the gun is \(4,244.4 \, \mathrm{N}\). This showcases how small objects can exert powerful impacts due to high acceleration, emphasizing the power of physics in real-world applications.

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Most popular questions from this chapter

How large a horizontal force in addition to \(F_{T}\) must pull on block- \(A\) in Fig. \(3-21\) to give it an acceleration of \(0.75 \mathrm{~m} / \mathrm{s}^{2}\) toward the left? Assume, as in Problem \(3.31\), that \(\mu_{k}=0.20, m_{A}=25 \mathrm{~kg}\), and \(m_{B}=15 \mathrm{~kg}\). Redraw Fig 3-21 for this case, including a force \(F\) pulling toward the left on \(A\). In addition, the retarding friction force \(F_{\mathrm{f}}\) must be reversed in direction. As in Problem 3.31, \(F_{\mathrm{f}}=49.1 \mathrm{~N}\). Write \(F=m a\) for each block in turn, taking the direction of motion (to the left and up) to be positive. We have $$ \begin{aligned} \pm \sum F_{x A}=F-F_{T}-49.1 \mathrm{~N} &=(25 \mathrm{~kg})\left(0.75 \mathrm{~m} / \mathrm{s}^{2}\right) \text { and }+\uparrow \sum F_{y B}=F_{T}-(15)(9.81) \mathrm{N} \\ &=(15 \mathrm{~kg})\left(0.75 \mathrm{~m} / \mathrm{s}^{2}\right) \end{aligned} $$ Solve the last equation for \(F_{T}\) and substitute in the previous equation. Then solve for the single unknown \(F\), and find it to be \(226 \mathrm{~N}\) or \(0.23 \mathrm{kN}\).

A child pulls on a rope attached to a sled with a force of \(60 \mathrm{~N}\). The rope makes an angle of \(40^{\circ}\) to the ground. ( \(a\) ) Compute the effective value of the pull tending to move the sled along the ground. (b) Compute the force tending to lift the sled vertically. As depicted in Fig. \(3-5\), the components of the \(60 \mathrm{~N}\) force are \(39 \mathrm{~N}\) and \(46 \mathrm{~N}\). ( \(a\) ) The pull along the ground is the horizontal component, \(46 \mathrm{~N}\). (b) The lifting force is the vertical component, \(39 \mathrm{~N}\).

The fabled planet Dune has a diameter eight times that of Earth and a mass twice as large. If a robot weighs \(1800 \mathrm{~N}\) on the surface of (nonspinning) Dune, what will it weigh at the poles on Earth? Take our planet to be a sphere.

A man who weighs \(1000 \mathrm{~N}\) on Earth stands on a scale on the surface of the mythical nonspinning planet Mongo. That body has a mass which is \(4.80\) times Earth's mass and a diameter which is \(0.500\) times Earth's diameter. Neglecting the effect of the Earth's spin, how much does the scale read?

(a) What is the smallest force parallel to a \(37^{\circ}\) incline needed to keep a \(100-\mathrm{N}\) weight from sliding down the incline if the coefficients of static and kinetic friction are both \(0.30 ?(b)\) What parallel force is required to keep the weight moving up the incline at constant speed? ( \(c\) ) If the parallel pushing force is \(94 \mathrm{~N}\), what will be the acceleration of the object? \((d)\) If the object in \((c)\) starts from rest, how far will it move in \(10 \mathrm{~s}\) ?

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