/*! 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

A 292 -kg motorcycle is accelerating up along a ramp that is inclined \(30.0^{\circ}\) above the horizontal. The propulsion force pushing the motorcycle up the ramp is \(3150 \mathrm{N}\), and air resistance produces a force of \(250 \mathrm{N}\) that opposes the motion. Find the magnitude of the motorcycle's acceleration.

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?

When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky diver and, thus, slows him down. Suppose the weight of the sky diver is \(915 \mathrm{N}\) and the drag force has a magnitude of \(1027 \mathrm{N}\). The mass of the sky diver is \(93.4 \mathrm{kg}\). What are the magnitude and direction of his acceleration?

A raindrop has a mass of \(5.2 \times 10^{-7} \mathrm{kg}\) and is falling near the surface of the earth. Calculate the magnitude of the gravitational force exerted (a) on the raindrop by the earth and (b) on the earth by the raindrop.

A block is pressed against a vertical wall by a force \(\overrightarrow{\mathbf{P}}\), as the drawing shows. This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity. The magnitude of the force is different in the two cases, while the directional angle \(\theta\) is the same. Kinetic friction exists between the block and the wall, and the coefficient of kinetic friction is \(0.250 .\) The weight of the block is \(39.0 \mathrm{N}\), and the directional angle for the force \(\overrightarrow{\mathbf{P}}\) is \(\theta=30.0^{\circ} .\) Determine the magnitude of \(\overrightarrow{\mathbf{P}}\) when the block slides (a) up the wall and (b) down the wall.

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