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Chapter 4: Problem 71

A \(15-g\) bullet is fired from a rifle. It takes \(2.50 \times 10^{-3}\) s for the bullet to travel the length of the barrel, and it exits the barrel with a speed of \(715 \mathrm{m} / \mathrm{s}\). Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet.

Short Answer

Expert verified
The average net force is 4,290 N.

Step by step solution

01

Determine the Bullet's Acceleration

First, use the kinematic equation for motion, which relates acceleration (#a), final velocity (#v), initial velocity (#v_0), and time (#t):\[v = v_0 + a \, t\]Given that the initial velocity #v_0 = 0 m/s (the bullet starts from rest), #v = 715 m/s, and #t = 2.50 \times 10^{-3} s. Solve for acceleration (#a):\[a = \frac{v - v_0}{t} = \frac{715 \text{ m/s} - 0 \text{ m/s}}{2.50 \times 10^{-3} \text{ s}}\]Calculate this to find the acceleration.
02

Calculate the Bullet's Acceleration

Perform the calculation of the acceleration:\[a = \frac{715}{2.50 \times 10^{-3}}\]which gives\[a = 286,000 \text{ m/s}^2\]This is the constant acceleration of the bullet as it travels through the barrel.
03

Use Newton's Second Law to Find the Force

Newton's second law states that the force (#F) is the product of mass (#m) and acceleration (#a):\[F = m \, a\]Convert the mass from grams to kilograms to use SI units (#15 \text{ g} = 0.015 \text{ kg}). Then, calculate the force:\[F = 0.015 \text{ kg} \times 286,000 \text{ m/s}^2\]Perform this calculation to obtain the average net force exerted on the bullet.
04

Calculate the Average Net Force

Carry out the multiplication to find the average net force:\[F = 0.015 \times 286,000 = 4,290 \text{ N}\]The average net force exerted on the bullet is 4,290 Newtons.

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Key Concepts

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

Kinematics
Understanding kinematics is fundamental to analyzing the motion of objects, like the bullet in our exercise. Kinematics focuses on motion without considering the forces causing it. Using kinematic equations, we can relate key variables such as velocity, acceleration, time, and displacement.

In this exercise, the bullet's journey through the barrel is described in kinematic terms. We start by using the equation of motion \[ v = v_0 + a \, t \]where:
  • \( v_0 \) is the initial velocity,
  • \( a \) is the acceleration,
  • \( t \) is the time taken.
Given that the initial velocity \( v_0 \) is zero (the bullet starts from rest), the bullet reaches a final speed \( v \) of 715 m/s in a short time span \( t \) of 2.50 x 10鈦宦� seconds. By rearranging the formula, we solve for the bullet's acceleration.
Constant Acceleration
The concept of constant acceleration simplifies the analysis of motion. If an object, such as our bullet, moves with constant acceleration, its speed increases at a uniform rate. This means its velocity changes linearly over time.

In our exercise, the bullet acquires high speed through the rifle barrel under constant acceleration. We calculated this using:\[ a = \frac{v - v_0}{t} \]Since the initial speed \( v_0 \) is zero, we can simplify to:\[ a = \frac{v}{t} \]Substituting the given values, we determine that the bullet's acceleration is an impressive 286,000 m/s虏. While real-world conditions might cause slight variations, assuming constant acceleration is useful for simplifying our estimate of the forces at play.
Average Net Force
Newton's Second Law connects force, mass, and acceleration. It states that the force acting on an object is the product of its mass and its acceleration.For our bullet, the average net force is computed as:\[ F = m \, a \]First, we converted the bullet's mass from grams to kilograms for consistency in SI units: \[ 15 \, \text{g} = 0.015 \, \text{kg} \]Then, we used the acceleration we previously calculated:\[ F = 0.015 \, \text{kg} \times 286,000 \, \text{m/s}^2 \]The resulting force is 4,290 Newtons. This force reflects the average push exerted by the rifle's gases on the bullet throughout the barrel. Understanding average net force helps us comprehend how objects like bullets can achieve such high speeds with short-duration forces.

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Most popular questions from this chapter

A billiard ball strikes and rebounds from the cushion of a pool table perpendicularly. The mass of the ball is 0.38 kg. The ball approaches the cushion with a velocity of \(+2.1 \mathrm{m} / \mathrm{s}\) and rebounds with a velocity of \(-2.0 \mathrm{m} / \mathrm{s} .\) The ball remains in contact with the cushion for a time of \(3.3 \times 10^{-3} \mathrm{s} .\) What is the average net force (magnitude and direction) exerted on the ball by the cushion?

In the amusement park ride known as Magic Mountain Superman, powerful magnets accelerate a car and its riders from rest to \(45 \mathrm{m} / \mathrm{s}\) (about \(100 \mathrm{mi} / \mathrm{h})\) in a time of \(7.0 \mathrm{s}\). The combined mass of the car and riders is \(5.5 \times 10^{3} \mathrm{kg} .\) Find the average net force exerted on the car and riders by the magnets.

A space traveler weighs \(540.0 \mathrm{N}\) on earth. What will the traveler weigh on another planet whose radius is twice that of earth and whose mass is three times that of earth?

A \(1.14 \times 10^{4}\) -kg lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is \(1.60 \mathrm{m} / \mathrm{s}^{2}\). At an altitude of \(165 \mathrm{m}\) the craft's downward velocity is \(18.0 \mathrm{m} / \mathrm{s}\). To slow down the craft, a retrorocket is firing to provide an upward thrust. Assuming the descent is vertical, find the magnitude of the thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface.

To hoist himself into a tree, a \(72.0-\mathrm{kg}\) man ties one end of a nylon rope around his waist and throws the other end over a branch of the tree. He then pulls downward on the free end of the rope with a force of \(358 \mathrm{N}\). Neglect any friction between the rope and the branch, and determine the man's upward acceleration.
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