/*! 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

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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 is approximately \( 37.33 \, \mathrm{N} \).

Step by step solution

01

Identify the Known Values

We need to identify the known variables in the problem. The mass of the fist, \( m = 0.70 \, \mathrm{kg} \), the final velocity, \( v = 8.0 \, \mathrm{m/s} \), the initial velocity, \( u = 0 \, \mathrm{m/s} \), and the time taken, \( t = 0.15 \, \mathrm{s} \).
02

Calculate the Acceleration

Use the formula for acceleration, \( a = \frac{v - u}{t} \). Substitute the known values: \( a = \frac{8.0 \, \mathrm{m/s} - 0 \, \mathrm{m/s}}{0.15 \, \mathrm{s}} \). This simplifies to \( a = \frac{8.0}{0.15} \), giving \( a \approx 53.33 \, \mathrm{m/s^2} \).
03

Apply Newton's Second Law

Newton's second law is given by \( F = ma \). We can use this to calculate the net force: \( F = 0.70 \, \mathrm{kg} \times 53.33 \, \mathrm{m/s^2} \). This results in \( F \approx 37.33 \, \mathrm{N} \).
04

Provide the Solution

The magnitude of the average net force applied to the fist is approximately \( 37.33 \, \mathrm{N} \).

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Key Concepts

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

Force Calculation
Force calculation is an essential concept in physics, particularly when dealing with motion and interaction between objects. According to Newton's Second Law, the force applied to an object is the product of its mass and the acceleration it experiences. This can be expressed as:\[ F = ma \]where:
  • \( F \) is the force in Newtons (N),
  • \( m \) denotes the mass in kilograms (kg),
  • and \( a \) is the acceleration in meters per second squared \((m/s^2)\).
When you know the mass of an object and its acceleration, you can find the force by multiplying these two quantities. In our example with the karate champion's fist, with a mass of \(0.70 \text{ kg}\) and an acceleration primary focus on deriving the value from given data, you can find that the applied force is about \(37.33 \text{ N}\). This simple law forms the basis for understanding how objects move and interact with forces in various mechanical systems.
Understanding force calculations is key to solving a wide range of problems in mechanics and beyond.
Acceleration Formula
The acceleration formula is pivotal in solving many mechanics problems as it helps us understand how quickly an object's velocity changes over time. It is defined by the equation:\[ a = \frac{v - u}{t} \]where:
  • \( a \) represents the acceleration,
  • \( v \) is the final velocity of the object,
  • \( u \) is the initial velocity,
  • and \( t \) is the time over which the change occurs.
In our example, the black belt’s fist started from rest, meaning the initial velocity \( u = 0 \text{ m/s} \).After \(0.15 \text{ s}\), it reached \(8.0 \text{ m/s}\).Substituting these values into the formula gives an acceleration of approximately \(53.33 \text{ m/s}^2\).Being familiar with this formula will empower you to solve a variety of problems where changes in motion are central.
It's a fundamental skill in physics that helps bridge concepts between dynamics and kinematics.
Mechanics Problem Solving
Mechanics problem solving involves applying physics concepts and mathematical tools to decipher real-world scenarios involving motion. At its core, problem-solving in mechanics requires:
  • Identifying known values and relevant physical laws,
  • Formulating equations based on these laws,
  • and systematically solving for the unknowns.
In the karate fist problem, we used known parameters like mass, velocity, and time to first compute acceleration; then applied Newton's Second Law to find the force. A streamlined approach involves:
  • Breaking down complex steps into smaller parts,
  • using logic and formulae fitting the scenario at hand,
  • and double-checking through dimensional analysis or units.
Mastering these techniques not only sharpens your analytical skills but also helps you tackle a wide array of problems efficiently. This systematic approach is essential for understanding the movement and interactions of objects in physical environments.

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Most popular questions from this chapter

The mass of a robot is 5450 kg. This robot weighs \(3620 \mathrm{N}\) more on planet A than it does on planet B. Both planets have the same radius of \(1.33 \times 10^{7} \mathrm{m} .\) What is the difference \(M_{\mathrm{A}}-M_{\mathrm{B}}\) in the masses of these planets?

A student is skateboarding down a ramp that is \(6.0 \mathrm{m}\) long and inclined at \(18^{\circ}\) with respect to the horizontal. The initial speed of the skateboarder at the top of the ramp is \(2.6 \mathrm{m} / \mathrm{s}\). Neglect friction and find the speed at the bottom of the ramp.

Two forces \(\overrightarrow{\mathbf{F}}_{\mathrm{A}}\) and \(\overrightarrow{\mathbf{F}}_{\mathrm{B}}\) are applied to an object whose mass is \(8.0 \mathrm{kg} .\) The larger force is \(\overrightarrow{\mathbf{F}}_{\mathrm{A}} .\) When both forces point due east, the object's acceleration has a magnitude of \(0.50 \mathrm{m} / \mathrm{s}^{2} .\) However, when \(\overrightarrow{\mathbf{F}}_{\mathrm{A}}\) points due east and \(\overrightarrow{\mathbf{F}}_{\mathrm{B}}\) points due west, the acceleration is \(0.40 \mathrm{m} / \mathrm{s}^{2}\), due east. Find (a) the magnitude of \(\overrightarrow{\mathbf{F}}_{A}\) and (b) the magnitude of \(\overrightarrow{\mathbf{F}}_{\mathrm{B}}\).

The alarm at a fire station rings and an \(86-\mathrm{kg}\) fireman, starting from rest, slides down a pole to the floor below (a distance of \(4.0 \mathrm{m}\) ). Just before landing, his speed is \(1.4 \mathrm{m} / \mathrm{s}\). What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?

During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between two fence posts. The limb exerts a downward force of \(151 \mathrm{N}\) on the wire. The left section of the wire makes an angle of \(14.0^{\circ}\) relative to the horizontal and sustains a tension of \(447 \mathrm{N} .\) Find the magnitude and direction of the tension that the right section of the wire sustains.

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