/*! 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 54 A \(12.0-\mathrm{g}\) bullet is ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(12.0-\mathrm{g}\) bullet is accelerated from rest to a speed of \(700 \mathrm{~m} / \mathrm{s}\) as it travels \(20.0 \mathrm{~cm}\) in a gun barrel. Assuming the acceleration to be constant, how large was the accelerating force? [Hint: Be careful of units.]

Short Answer

Expert verified
The accelerating force was 14,700 N.

Step by step solution

01

Understand the Problem

We need to find the accelerating force that was applied to a bullet. We know the mass of the bullet as well as its initial and final velocities, and the distance it traveled to reach that speed.
02

Convert Units

Convert all units to standard SI units. The mass is given as \(12.0\) grams, which converts to \(0.012\) kg, and the distance is given as \(20.0\) cm, which converts to \(0.20\) meters.
03

Identify Known Variables

The known values are: mass \(m = 0.012\) kg, initial velocity \(v_i = 0\) m/s, final velocity \(v_f = 700\) m/s, and distance \(d = 0.20\) m. We need to find the force \(F\).
04

Use Kinematic Equation to Find Acceleration

Since acceleration is constant, use the kinematic equation: \(v_f^2 = v_i^2 + 2a d\). Solve for \(a\): \(a = \frac{v_f^2 - v_i^2}{2d}\).
05

Calculate Acceleration

Substitute the given values into the kinematic equation: \(a = \frac{700^2 - 0^2}{2 \times 0.20}\). Calculate \(a\).
06

Use Newton's Second Law to Calculate Force

Using Newton's second law, \(F = ma\), substitute \(m\) and \(a\) to find the force.
07

Final Calculation

From Step 5, \(a = 1,225,000\) m/s². Using \(F = 0.012 \times 1,225,000\), calculate \(F\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Newton's Second Law
Newton's Second Law of Motion provides us with a fundamental relation in physics that links force, mass, and acceleration. The law is given by the equation \( F = ma \), where \( F \) is the force applied to an object, \( m \) is the mass of the object, and \( a \) is the acceleration produced by the force. This means that the force acting on an object is equal to the mass of that object multiplied by the acceleration it undergoes. Using this law in the bullet problem, once we find the acceleration, we can easily determine the force needed to achieve that acceleration by multiplying it by the bullet's mass. It's essential to use consistent units—typically the International System of Units (SI)—to ensure correctly calculated results.
Kinematic Equations
Kinematic equations relate the variables of motion, such as velocity, acceleration, distance, and time. These equations are pivotal for solving problems involving constant acceleration, as seen in our bullet problem. The specific kinematic equation used here is \( v_f^2 = v_i^2 + 2ad \), where:
  • \( v_f \) is the final velocity
  • \( v_i \) is the initial velocity
  • \( a \) is the acceleration
  • \( d \) is the distance traveled
This equation helps us solve for acceleration when we know the initial and final velocities and the distance over which the acceleration occurs. It is a particularly useful tool when time is not a factor in the calculations.
Unit Conversion
Unit conversion is critical, as it allows for consistent and accurate computations. In physics, using the correct units is crucial for ensuring that equations and calculations are dimensionally accurate. Typically, we convert all quantities into SI units before plugging them into equations. For our bullet problem:
  • The bullet's mass was given in grams, which was converted to kilograms for standard SI use (\( 12.0\,g = 0.012\,kg \)).
  • The distance was given in centimeters, which was converted to meters (\( 20.0\,cm = 0.20\,m \)).
Without converting these units, the final calculation of force using Newton’s Second Law would be inaccurate.
Constant Acceleration
Constant acceleration implies that the acceleration remains the same over the distance or time considered. This assumption simplifies the use of kinematic equations, making them direct and applicable for solving motion problems without factoring in changing forces or velocities. In the given problem, we are dealing with constant acceleration from the bullet's state when it is at rest, to its high-speed motion within the short span of the gun barrel. The constant acceleration confirms that the acceleration value calculated can be used directly in subsequent calculations, such as when applying Newton's Second Law. This concept is particularly important in practical scenarios where forces do not fluctuate rapidly and can be considered stable over the period of interest.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the weight on the surface of the Earth of a body whose mass is (a) \(3.00 \mathrm{~kg}\), and \((b) 200 \mathrm{~g}\). The general relation between mass \(m\) and weight \(F_{W}\) is \(F_{W}=m g\). In this relation, \(m\) must be in kilograms, \(g\) in meters per second squared, and \(F_{W}\) in newtons. On Earth, \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\). The acceleration due to gravity varies from place to place in the universe. (a) \(\quad F_{W}=(3.00 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=29.4 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}=29.4 \mathrm{~N}\) (b) \(F_{W}=(0.200 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=1.96 \mathrm{~N}\)

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?

The Earth's radius is about \(6370 \mathrm{~km}\). An object that has a mass of \(20 \mathrm{~kg}\) is taken to a height of \(160 \mathrm{~km}\) above the Earth's surface. ( \(a\) ) What is the object's mass at this height? ( \(b\) ) How much does the object weigh (i.e. how large a gravitational force does it experience) at this height?

(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 \(10 \mathrm{~s}\) ?

A force of \(100 \mathrm{~N}\) makes an angle of \(\theta\) with the \(x\) -axis and has a scalar \(y\) -component of \(30 \mathrm{~N}\). Find both the scalar \(x\) -component of the force and the angle \(\theta\). (Remember that the number \(100 \mathrm{~N}\) has three significant figures whereas \(30 \mathrm{~N}\) has only two.)Begin your analysis by drawing a diagram. Here the data are sketched roughly in Fig. \(3-4\). We wish to find \(F_{x}\) and \(\theta\). Since $$ \sin \theta=\frac{30 \mathrm{~N}}{100 \mathrm{~N}}=0.30 $$ \(\theta=17.46^{\circ}\), and thus, to two significant figures, \(\theta=17^{\circ} .\) Then, using the \(\cos \theta\) $$ F_{x}=(100 \mathrm{~N}) \cos 17.46^{\circ}=95 \mathrm{~N} $$

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.