/*! 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 65 In Galileo drops a 2.5 -kg canno... [FREE SOLUTION] | 91Ó°ÊÓ

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In Galileo drops a 2.5 -kg cannon ball from the 58.4 -m Tower of Pisa. If the ball makes a \(0.130-\mathrm{m}\) -deep hole in the ground, find the average force exerted by the ground on the ball.

Short Answer

Expert verified
The average force exerted by the ground on the ball is approximately 11006.31 N.

Step by step solution

01

Understand the Problem Context

We need to find the average force exerted by the ground on the ball when it impacts the ground after being dropped from a height. We have the initial height from which the ball is dropped, the depth of the hole made by the ball, and the mass of the ball.
02

Apply the Conservation of Energy

When the ball is dropped, it has potential energy which gets converted into kinetic energy just before it hits the ground. Calculate the initial potential energy using \[ PE_{initial} = mgh \]where \( m = 2.5 \text{ kg} \) (mass of the ball), \( g = 9.81 \text{ m/s}^2 \) (acceleration due to gravity), \( h = 58.4 \text{ m} \) (height).Plug in the values:\[ PE_{initial} = 2.5 \times 9.81 \times 58.4 \]
03

Calculate Kinetic Energy on Impact

Since all the potential energy is converted to kinetic energy upon impact, the kinetic energy at the moment of impact is the same as the initial potential energy:\[ KE_{impact} = PE_{initial} \]
04

Use Work-Energy Principle

When the ball hits the ground, it comes to a stop after making a hole. The work done by the ground to stop the ball is equal to the kinetic energy the ball had at the point of impact. Work done is defined as the force times the distance over which it acts, thus:\[ F \cdot d = KE_{impact} \]Solve for the force \( F \):\[ F = \frac{KE_{impact}}{d} \]where \( d = 0.130 \text{ m} \) is the depth of the hole.
05

Calculate the Average Force

First, calculate the initial potential energy:\[ PE_{initial} = 2.5 \times 9.81 \times 58.4 = 1430.82 \text{ J} \]This is equal to the kinetic energy at impact. Now, find the average force:\[ F = \frac{1430.82}{0.130} = 11006.31 \text{ N} \]

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Key Concepts

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

Conservation of Energy
When thinking about physics, the Conservation of Energy principle is a very useful tool. It states that energy cannot be created or destroyed; it can only change forms. This means that the total amount of energy in a closed system remains constant.

For instance, in our problem about the cannonball, we see this principle clearly. Initially, the cannonball has potential energy because it is elevated off the ground. This potential energy is calculated by the formula \( PE = mgh \), where \( m \) is the mass (2.5 kg), \( g \) is the acceleration due to gravity (9.81 m/s²), and \( h \) is the height (58.4 m).

As the cannonball falls, it loses height, so its potential energy decreases. However, as the potential energy decreases, its velocity increases, turning that lost potential energy into kinetic energy just before impact. In essence, all the potential energy initially present is transformed into kinetic energy by the time it reaches the ground.
Kinetic Energy
Kinetic Energy is the energy that an object possesses due to its motion. When the cannonball falls, its velocity increases, and therefore its kinetic energy increases.

The formula for Kinetic Energy is \( KE = \frac{1}{2}mv^2 \), which in simpler terms means that kinetic energy is half of the mass times the velocity squared. However, in our case, we don’t need to calculate the velocity; instead, we directly equate the initial potential energy to the kinetic energy just before the impact.

This transition from potential to kinetic energy highlights how energy moves from one form to another. By the time the cannonball hits the ground, all its potential energy has become kinetic energy, showing that energy conversion is fundamental in physics problems.
Work-Energy Principle
The Work-Energy Principle is another important concept that plays a significant role in understanding the forces involved when the cannonball hits and penetrates the ground. According to this principle, the work done on an object results in a change in its kinetic energy.

In this scenario, when the cannonball hits the ground, its kinetic energy is used to do work to make a hole in the ground. The relation is given by \( W = F \cdot d \), where \( W \) is the work done (which equals the kinetic energy at impact), \( F \) is the force exerted by the ground, and \( d \) is the distance over which the force acts (depth of the hole, 0.130 m).

Solving for the force, we use the equation \( F = \frac{W}{d} \). This approach helps us understand how much force was necessary through the action of the cannonball digging into the soil, demonstrating how energy transformation plays into the force calculations.

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Most popular questions from this chapter

You throw a tennis ball straight up, and it peaks at height \(H\) above the launch point. Ignore air resistance. (a) For each of the following points, draw vectors indicating the net force acting on the ball: (i) just after release; (ii) going up through \(H / 2\); (iii) at \(H,\) the peak; (iv) coming down through \(H / 2 ;\) (v) coming down through the launch point. (b) For each of the five points, draw the ball's velocity vector.

In A man pushes a 32 -kg lawn mower using a handle inclined \(40^{\circ}\) from the horizontal. He pushes with \(65 \mathrm{~N}\) directed along the handle. (a) What's the mower's weight? (b) What's the normal force on the mower? (c) What's the mower's acceleration (ignoring friction)?

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