/*! 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 96 A person with a black belt in ka... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A person with a black belt in karate has a fist that has a mass of 0.70 \(\mathrm{kg}\) . Starting from rest, this fist attains a velocity of 8.0 \(\mathrm{m} / \mathrm{s}\) in 0.15 \(\mathrm{s}\) . What is the magnitude of the average net force applied to the fist to achieve this level of performance?

Short Answer

Expert verified
The average net force applied is approximately 37.33 N.

Step by step solution

01

Identifying Known Values

To find the average net force, we first identify the values provided in the problem. The mass (\(m\)) of the fist is 0.70 kg, the initial velocity (\(u\)) is 0 m/s, the final velocity (\(v\)) is 8.0 m/s, and the time interval (\(t\)) is 0.15 s.
02

Calculating Acceleration

To calculate the acceleration, we use the formula: \(a = \frac{v - u}{t}\). Substituting the values, we have \(a = \frac{8.0 \, \mathrm{m/s} - 0 \, \mathrm{m/s}}{0.15 \, \mathrm{s}} = \frac{8.0}{0.15} \, \mathrm{m/s^2}\). Calculating this gives \(a \approx 53.33 \, \mathrm{m/s^2}\).
03

Using Newton's Second Law

Newton's second law states that the force is equal to the mass times the acceleration \(F = ma\). Substituting in our mass (0.70 kg) and the acceleration (53.33 \(\mathrm{m/s^2}\)), we find \(F = 0.70 \, \mathrm{kg} \times 53.33 \, \mathrm{m/s^2} = 37.33 \, \mathrm{N}\).
04

Rounding the Force

For simplicity and significance, round the value of force to two decimal places, which results in \(F \approx 37.33 \, \mathrm{N}\).

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.

Acceleration Calculation
Calculating acceleration is an essential step in understanding how fast an object's velocity is changing over time. In this scenario, where a karate expert's fist goes from rest to a speed of 8.0 m/s within 0.15 seconds, we're tasked with finding that rate of change - the acceleration.
To calculate acceleration, use the formula:
  • \[ a = \frac{v - u}{t} \]
Here, "\(v\)" represents the final velocity, "\(u\)" is the initial velocity, and "\(t\)" is the time period over which this change occurs.
Initially, the fist is at rest, making the initial velocity zero. Substituting our values gives us:
  • \[ a = \frac{8.0 \, \text{m/s} - 0 \, \text{m/s}}{0.15 \, \text{s}} \]
  • \[ a \approx 53.33 \, \text{m/s}^2 \]
This means the fist's speed is increasing by about 53.33 meters per second every second.
Newton's Second Law
Newton's Second Law of Motion is a cornerstone of classical mechanics. It tells us about the relationship between an object's mass, its acceleration, and the force applied to it.
According to this law:
  • The force acting on an object is equal to the mass of that object multiplied by its acceleration.
  • This can be expressed as \( F = ma \).
In our exercise, we already calculated the acceleration of the karate expert’s fist to be 53.33 m/s², and we know the mass of the fist is 0.70 kg.
Now, substituting these values into Newton's Second Law yields:
  • \[ F = 0.70 \, \text{kg} \times 53.33 \, \text{m/s}^2 \]
  • \[ F \approx 37.33 \, \text{N} \]
Thus, 37.33 Newtons is the force required to accelerate the fist.
Mass and Velocity
Mass and velocity are two key players in understanding motion and its effects.
- **Mass (m):** This is how much matter an object contains. In our scenario, the karate expert's fist has a mass of 0.70 kg. Mass doesn't change regardless of the object's motion. However, it significantly influences how the object responds to forces. - **Velocity (v):** This describes how fast something is moving and in what direction. Starting from rest, the karate expert's fist achieves a velocity of 8.0 m/s. Initial velocity is zero if the object begins from a complete stop. The combination of mass and velocity measurements helps us discover other properties of motion, like momentum, which plays a crucial role in understanding the effects of forces applied during motion.
Time Interval in Motion
The time interval in motion refers to the period during which changes in an object's position or velocity are observed. It's crucial for calculating things like acceleration.
In the karate example, the fist reaches a velocity of 8.0 m/s from rest over a time interval of 0.15 seconds. The time interval is a key component because:

  • It allows us to determine how quickly the fist is accelerating. Shorter time intervals for the same velocity change mean higher acceleration.

  • This helps calculate how much force is applied to the fist, since force is influenced by both mass and acceleration.
Understanding time intervals provides insights into the dynamics of motion and how different variables interplay in the laws of physics.

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

On earth, two parts of a space probe weigh 11000 \(\mathrm{N}\) and 3400 \(\mathrm{N}\) . These parts are separated by a center-to-center distance of 12 \(\mathrm{m}\) and may be treated as uniform spherical objects. Find the magnitude of the gravitational force that each part exerts on the other out in space, far from any other objects.

mmh The drawing shows a large cube (mass \(=25 \mathrm{kg}\) ) being accelerated across a horizontal frictionless surface by a horizontal force \(\overrightarrow{\mathbf{P}}\) . A small cube (mass \(=4.0 \mathrm{kg} )\) is in contact with the front surface of the large cube and will slide downward unless \(\overrightarrow{\mathbf{P}}\) is sufficiently large. The coefficient of static friction between the cubes is \(0.71 .\) What is the smallest magnitude that \(\overrightarrow{\mathbf{P}}\) can have in order to keep the small cube from sliding downward?

ssm A 60.0-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.760 and 0.410, respectively. What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at a constant speed?

A space probe has two engines. Each generates the same amount of force when fired, and the directions of these forces can be independently adjusted. When the engines are fired simultaneously and each applies its force in the same direction, the probe, starting from rest, takes \(28 s\) to travel a certain distance. How long does it take to travel the same distance, again starting from rest, if the engines are fired simultaneously and the forces that they apply to the probe are perpendicular?

A damp washcloth is hung over the edge of a table to dry. Thus, part (mass \(=m_{\mathrm{ca}} )\) of the washcloth rests on the table and part (mass \(=m_{\text { oft }} )\) does not. The coefficient of static friction between the table and the washcloth is 0.40 . Determine the maximum fraction \(\left[m_{\mathrm{oft}} /\left(m_{\mathrm{cn}}+m_{\mathrm{off}}\right)\right]\) that can hang over the edge without causing the whole washcloth to slide off the table.

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.