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An object whose mass is \(0.092 \mathrm{~kg}\) is initially at rest and then attains a speed of \(75.0 \mathrm{~m} / \mathrm{s}\) in \(0.028 \mathrm{~s}\). What average net force acted on the object during this time interval? a) \(1.2 \cdot 10^{2} \mathrm{~N}\) b) \(2.5 \cdot 10^{2} \mathrm{~N}\) c) \(2.8 \cdot 10^{2} \mathrm{~N}\) d) \(4.9 \cdot 10^{2} \mathrm{~N}\)

Short Answer

Expert verified
a) \(2.0 \cdot 10^{2} \mathrm{~N}\) b) \(2.5 \cdot 10^{2} \mathrm{~N}\) c) \(3.0 \cdot 10^{2} \mathrm{~N}\) d) \(3.5 \cdot 10^{2} \mathrm{~N}\) Answer: b) \(2.5 \cdot 10^{2} \mathrm{~N}\)

Step by step solution

01

Identify the known variables

We are given the following information: - Mass (m) = 0.092 kg - Initial speed (v0) = 0 m/s - Final speed (vf) = 75.0 m/s - Time interval (t) = 0.028 s
02

Calculate the acceleration (a)

We can find the acceleration by using the kinematic equation: a = (vf - v0) / t Plugging in the known values, we get: a = (75.0 m/s - 0 m/s) / 0.028 s = (75 m/s) / 0.028 s ≈ 2678.57 \mathrm{~m/s^2}
03

Calculate the average net force (F)

Now that we have the acceleration, we can use Newton's second law to find the average net force acting on the object: F = m * a Plugging in the known values, we get: F = 0.092 kg * 2678.57 \mathrm{~m/s^2} ≈ 246.35 N
04

Choose the closest answer

Since 246.35 N is not exactly equal to any of the answer choices, we need to choose the closest one. The closest answer to 246.35 N is option b) \(2.5 \cdot 10^{2} \mathrm{~N}\).

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Key Concepts

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

Average Net Force
Understanding the average net force is essential when delving into the dynamics of an object's motion. In our daily experiences, force is a push or pull on an object that can cause it to accelerate, such as a person pushing a shopping cart. However, in physics, it's defined precisely and calculated using Newton's second law of motion. This fundamental principle states that the force applied to an object is directly proportional to the acceleration it experiences and is given by the equation \( F = ma \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration.

To determine the amount of force during a time interval where the velocity of an object changes, one calculates the average net force. This encompasses all the individual forces acting on the object and considers their sum as a single force causing the acceleration. It's crucial to remember that forces in opposite directions must be taken into account with their appropriate signs; when they are netted out, we get the 'net force' responsible for the change in motion.

In the case of the textbook exercise, the net force was found by first calculating the acceleration, using the given mass of the object and then applying Newton's second law. The calculation revealed that the average net force acting on a small 0.092 kg object to reach a speed of 75.0 m/s in just 0.028 seconds was roughly 246.35 N, which when rounded to the given answer choices, became 250 N.
Acceleration
Acceleration is a measure of how quickly the velocity of an object is changing. It is a vector quantity, which means it has both magnitude and direction. In simple terms, acceleration occurs whenever an object speeds up, slows down, or changes direction. Mathematically, it is calculated by the change in velocity over time, represented as \( a = \frac{\Delta v}{\Delta t} \), where \( \Delta v \) is the change in velocity and \( \Delta t \) is the time over which the change occurs.

Calculating Acceleration in our Example

In our original textbook problem, the object accelerates from 0 to 75.0 m/s over a period of 0.028 seconds. We can compute this acceleration by taking the final velocity and subtracting the initial velocity, then dividing by the time taken. The sheer size of the resulting acceleration, 2678.57 m/s², indicates a very rapid change in velocity—an important clue to the magnitude of forces at play. Students might be surprised by the high value of acceleration; emphasizing the brief time interval is key in explaining why such a high value is plausible. Indeed, short bursts of time can lead to significant accelerations.
Kinematics
Kinematics is the study of motion without considering the forces that cause it. It describes motion using concepts such as distance, displacement, speed, velocity, and acceleration. One of the most important aspects of kinematics is the set of equations that relate these quantities to each other and allow us to solve motion problems.

In kinematics, an essential quantity to understand is velocity, which is the rate of change of an object's position. Unlike speed, velocity includes information about the direction of movement. The example from the textbook solutions section consists of motion that can be classified as one-dimensional kinematics since it occurs in a straight line, which simplifies the calculations.

Kinematics in Solving Our Problem

To solve the example problem, we rely on kinematics to calculate the acceleration by knowing the initial and final velocities and the time interval. The step by step solution shows this process: the object’s initial velocity was zero, and it attained a velocity of 75.0 m/s over a time of 0.028 s. Kinematics provides the framework for us to understand and quantify how the object got from one state of motion to another, paving the way for using Newton's second law to find the average net force.

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Most popular questions from this chapter

Two blocks are in contact on a frictionless, horizontal tabletop. An external force, \(\vec{F},\) is applied to block \(1,\) and the two blocks are moving with a constant acceleration of \(2.45 \mathrm{~m} / \mathrm{s}^{2} .\) Use \(M_{1}=3.20 \mathrm{~kg}\) and \(M_{2}=5.70 \mathrm{~kg}\) a) What is the magnitude, \(F\), of the applied force? b) What is the contact force between the blocks? c) What is the net force acting on block \(1 ?\)

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