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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 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 \(F \approx 37.33 \, \mathrm{N}\).

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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.

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Most popular questions from this chapter

A stuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. The cable is parallel to the ground. The mass of the stuntman is 109 kg, and the coefficient of kinetic friction between the road and him is 0.870. Find the tension in the cable.

The principles used to solve this problem are similar to those in Multiple- Concept Example 17. A 205-kg log is pulled up a ramp by means of a rope that is parallel to the surface of the ramp. The ramp is inclined at 30.0 with respect to the horizontal. The coefficient of kinetic friction between the log and the ramp is 0.900, and the log has an acceleration of magnitude 0.800 \(\mathrm{m} / \mathrm{s}^{2}\) . Find the tension in the rope.

As part \(a\) of the drawing shows, two blocks are connected by a rope that passes over a set of pulleys. One block has a weight of 412 \(\mathrm{N}\) , and the other has a weight of 908 \(\mathrm{N}\) . The rope and the pulleys are mass-less and there is no friction. (a) What is the acceleration of the lighter block? (b) Suppose that the heavier block is removed, and a downward force of 908 \(\mathrm{N}\) is provided by someone pulling on the rope, as part \(b\) of the drawing shows. Find the acceleration of the remaining block. (c) Explain why the answers in (a) and (b) are different.

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.

Consult Multiple-Concept Example 10 in preparation for this problem. Traveling at a specd of \(16.1 \mathrm{m} / \mathrm{s},\) the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is \(0.720 .\) What is the speed of the automobile after 1.30 \(\mathrm{s}\) have clapsed? Ignore the effects of air resistance.

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