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Chapter 3: Problem 51

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 \(16.2\) -g bullet within the gun, and then compute the average force exerted on it.

Short Answer

Expert verified
Average acceleration: \(262,000\,\text{m/s}^2\). Average force: \(4,244.4\,\text{N}\).

Step by step solution

01

Understand Given Data

We're given the speed of the bullet as it leaves the barrel: \( v = 262 \text{ m/s} \). The time it takes to traverse the barrel \( t = 1 \text{ ms} = 0.001 \text{ s} \). The mass of the bullet is \( m = 16.2 \text{ g} = 0.0162 \text{ kg} \).
02

Calculate Average Acceleration

Use the formula for acceleration \( a = \frac{\Delta v}{\Delta t} \), where the change in velocity \( \Delta v = v = 262 \text{ m/s} \) since initial speed is 0. Plugging in the values, we get:\[ a = \frac{262 \text{ m/s}}{0.001 \text{ s}} = 262,000 \text{ m/s}^2 \].
03

Calculate Average Force

The average force \( F \) is found using Newton's second law \( F = ma \). Using the mass \( m = 0.0162 \text{ kg} \) and the acceleration \( a = 262,000 \text{ m/s}^2 \), we calculate:\[ F = 0.0162 \times 262,000 = 4,244.4 \text{ N} \].

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Key Concepts

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

Understanding Average Acceleration
Average acceleration is a measure of how quickly an object changes its velocity over a certain amount of time. To better understand this concept, imagine starting from rest and quickly speeding up to a high velocity. This rapid change in velocity is what we call acceleration.
The formula for calculating average acceleration (\( a \)) is given by:\[ a = \frac{\Delta v}{\Delta t} \] Where:
  • \( \Delta v \) is the change in velocity, and
  • \( \Delta t \) is the change in time.
In our bullet example, it started from a stationary position and reached a speed of \( 262 \ \text{m/s} \) in just \( 0.001 \ \text{s} \). Therefore, the full change in velocity is \( 262 \ \text{m/s} \) and the time taken is \( 0.001 \ \text{s} \). Using the formula, we calculate the average acceleration as \( 262,000 \ \text{m/s}^2 \). This value shows a very rapid increase in speed, which is typical for bullets as they exit a gun barrel.
Understanding average acceleration helps us to comprehend how forces can rapidly change an object's speed.
Calculating Average Force
Calculating the average force exerted on an object involves understanding the relationship between force, mass, and acceleration. A core principle in physics, especially when discussing movement and interactions, is that force causes changes in an object鈥檚 motion.
The formula to find average force (\( F \) ) is described by Newton's second law of motion:\[ F = ma\] In this equation:
  • \( F \) is the force,
  • \( m \) is the mass of the object, and
  • \( a \) is the acceleration.
Given our bullet with a mass of \( 0.0162 \ \text{kg} \) and an acceleration of \( 262,000 \ \text{m/s}^2 \), the average force exerted on the bullet is calculated as follows:\[ F = 0.0162 \times 262,000 = 4,244.4 \ \text{N}\] This impressive force value explains how bullets are propelled at such high speeds. It also exemplifies how even small objects can be given large kinetic energies when substantial forces are applied rapidly.
Explaining Newton's Second Law
Newton's second law is a fundamental tenet in physics that describes the relationship between an object鈥檚 mass, the acceleration it experiences, and the resultant force exerted. This law provides the formula for force as:\[ F = ma\]Where:
  • \( F \) is the force applied to the object,
  • \( m \) is the mass, and
  • \( a \) is the acceleration the object undergoes.
This law implies that the force needed to accelerate an object is proportional to the mass of the object and the acceleration. Therefore, for the same amount of force, a lighter object will accelerate more than a heavier one.
In the scenario with the bullet, Newton鈥檚 second law helps us understand why a small object like a bullet can be accelerated to such high speeds despite its lower mass. By applying a substantial force over a very short interval, the bullet achieves high velocity almost instantly. This relationship between force, mass, and acceleration is crucial in understanding motion dynamics across many real-world applications.

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Most popular questions from this chapter

A \(20.0 \mathrm{~kg}\) object that can move freely is subjected to a resultant force of \(45.0 \mathrm{~N}\) in the \(-x\) -direction. Find the acceleration of the object. We make use of the second law in component form, \(\Sigma F_{x}=m a_{x}\), with \(\Sigma F_{x}=-45.0 \mathrm{~N}\) and \(m=20.0 \mathrm{~kg}\). Then $$ a_{x}=\frac{\sum F_{x}}{m}=\frac{-45.0 \mathrm{~N}}{20.0 \mathrm{~kg}}=-2.25 \mathrm{~N} / \mathrm{kg}=-2.25 \mathrm{~m} / \mathrm{s}^{2} $$ where we have used the fact that \(1 \mathrm{~N}=1 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}\). Because the resultant force on the object is in the \(-x\) -direction, its acceleration is also in that direction.

A \(5.0\) -kg object is to be given an upward acceleration of \(0.30 \mathrm{~m} / \mathrm{s}^{2}\) by a rope pulling straight upward on it. What must be the tension in the rope? The free-body diagram for the object is shown in Fig. \(3-8(b)\). The tension in the rope is \(F_{T}\), and the weight of the object is \(F_{W}=m g=\) \((5.0 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=49.1 \mathrm{~N}\). Using \(\Sigma F_{y}=m a_{y}\) with up taken as positive, from which \(F_{T}=50.6 \mathrm{~N}=51 \mathrm{~N}\). As a check, we notice that \(F_{T}\) is larger than \(F_{W}\), as it must be if the object is to accelerate upward.

Once ignited, a small rocket motor on a spacecraft exerts a constant force of \(10 \mathrm{~N}\) for \(7.80 \mathrm{~s}\). During the burn, the rocket causes the 100 -kg craft to accelerate uniformly. Determine that acceleration.

A 300 -g mass hangs at the end of a string. A second string hangs from the bottom of that mass and supports a 900 -g mass. \((a)\) Find the tension in each string when the masses are accelerating upward at \(0.700 \mathrm{~m} / \mathrm{s}^{2}\). Don't forget gravity. ( \(b\) ) Find the tension in each string when the acceleration is \(0.700 \mathrm{~m} / \mathrm{s}^{2}\) downward.

The coefficient of static friction between a box and the flat bed of a truck is 0.60. What is the maximum acceleration the truck can have along level ground if the box is not to slide? The box experiences only one \(x\) -directed force, the friction force. When the box is on the verge of slipping, \(F_{f}=\mu \mathrm{s} F_{W}\), where \(F_{W}\) is the weight of the box. As the truck accelerates, the friction force must cause the box to have the same acceleration as the truck; otherwise, the box will slip. When the box is not slipping, \(\sum F_{x}=m a_{x}\) applied to the box gives \(F_{f}=m a_{x}\). However, if the box is on the verge of slipping, \(F_{f}\) \(=\mu_{s} F_{W}\) so that \(\mu_{s} F_{W}=m a_{x} .\) Because \(F_{W}=m g\), $$ a_{x}=\frac{\mu_{s} m g}{m}=\mu_{s} g=(0.60)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=5.9 \mathrm{~m} / \mathrm{s}^{2} $$ as the maximum acceleration without slipping.
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