/*! 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 23 A 2 \(300-\mathrm{kg}\) pile dri... [FREE SOLUTION] | 91Ó°ÊÓ

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A 2 \(300-\mathrm{kg}\) pile driver is used to drive a steel beam into the ground. The pile driver falls \(7.50 \mathrm{~m}\) before coming into contact with the top of the beam, and it drives the beam \(18.0 \mathrm{~cm}\) farther into the ground as it comes to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.

Short Answer

Expert verified
The average force the beam exerts on the pile driver while the pile driver is brought to rest is 122,625 N.

Step by step solution

01

Determine the initial potential energy

The potential energy of the pile driver before it falls is given by the equation \( P.E = mgh \), where m is the mass of the pile driver (300 kg), g is the acceleration due to gravity (9.81 m/s² ), and h is the height fallen (7.50 m). So, calculate it: \( P.E = 300 kg \cdot 9.81 m/s² \cdot 7.50 m = 22,072.5 \, J \).
02

Determine the final kinetic energy

At the point just before the pile driver hits the beam, all the potential energy has been converted to kinetic energy. So, the kinetic energy is same as the initial potential energy which is \( K.E = 22,072.5 \, J \).
03

Calculate the work done to stop the pile driver

The work done by the beam to stop the pile driver is equal to the kinetic energy of the pile driver. Hence, the work done is \( W = 22,072.5 \, J \).
04

Calculate the force exerted by the beam

The work done in stopping the pile driver over a distance of 0.18 m is given by \( W = Fd \), where F is the force exerted by the beam and d the distance moved (18.0 cm = 0.18 m). By rearranging the formula as \( F = W / d \), we can obtain the force: \( F = 22,072.5J / 0.18m = 122,625N \).

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Key Concepts

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

Potential Energy
Potential energy is the stored energy of an object due to its position relative to the ground. In physics, it's the energy that a body possesses because of its height. For a pile driver, this energy is accumulated during its ascent to a certain height from which it will later fall. The potential energy is determined using the formula \( P.E = mgh \), where \( m \) stands for mass, \( g \) is the acceleration due to gravity, and \( h \) is the height above a reference point.

In the context of our exercise, as the 300 kg pile driver is lifted to 7.5 meters, it gains potential energy precisely because of this elevation. As it stands atop this height, the pile driver holds within it an energy of 22,072.5 Joules, ready to be released as it falls.
  • This energy is directly dependent on the mass and height of the pile driver.
  • The gravitational pull (\( g \) is typically 9.81 m/s²) plays a crucial role in determining how much potential energy is stored.
Potential energy transforms into kinetic energy once the pile driver begins its descent.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. When the pile driver falls, its potential energy starts transforming into kinetic energy. By the time it reaches the top of the beam, all its potential energy has been converted into kinetic energy.

The formula for kinetic energy is \( K.E = \frac{1}{2}mv^2 \). However, in this scenario, we can equate the kinetic energy directly to the potential energy it had at the top because it converts all of it as it falls. Here, the kinetic energy becomes 22,072.5 Joules right before it impacts the beam.
Some important aspects to remember:
  • Kinetic energy increases with the velocity of the object.
  • At the highest point, kinetic energy is zero due to lack of motion.
  • At the lowest point (just before impact), potential energy is zero and kinetic energy is at its maximum.
The transformation between potential and kinetic energies is a hallmark of the Work-Energy Principle.
Average Force
Average force is a concept derived from the Work-Energy Principle, which states that the work done on an object is equal to its change in kinetic energy. When the pile driver hits the beam, it is brought to a stop over a certain distance.

To determine the average force exerted by the beam, we use the work-energy equation \( W = Fd \), where \( W \) is the work done, \( F \) is the force, and \( d \) is the distance over which the force is applied. From the solution, the work done is equal to the kinetic energy of 22,072.5 Joules, and the distance the beam moves is 0.18 meters.

We then solve for the force using \( F = \frac{W}{d} \), resulting in an astonishing average force of 122,625 Newtons applied by the beam on the pile driver. This immense force is responsible for bringing the descending pile driver to a halt over a very short distance, highlighting the powerful nature of energy interactions.
  • Average force calculates the total impact force distributed over the stopping distance.
  • A shorter stopping distance means a larger force, as seen in this exercise.
  • Understanding average force helps in engineering scenarios like designing safe equipment to absorb high impact forces efficiently.
In conclusion, leveraging the concepts of potential and kinetic energy alongside average force allows us to uncover surprising details about energy transformations during motion.

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Most popular questions from this chapter

A child's pogo stick (Fig. P5.81) stores energy in a spring \(\left(k=2.50 \times 10^{4} \mathrm{~N} / \mathrm{m}\right) .\) At position \((\mathrm{A})\left(x_{1}=-0.100 \mathrm{~m}\right)\) the spring compression is a maximum and the child is momentarily at rest. At position (B) \((x=0)\), the spring is relaxed and the child is moving upward. At position (C). the child is again momentarily at rest at the top of the jump. Assuming that the combined mass of child and pogo stick is \(25.0 \mathrm{~kg}\), (a) calculate the total energy of the system if both potential energies are zero at \(x=0\), (b) determine \(x_{2}\), (c) calculate the speed of the child at \(x=0,(\mathrm{~d})\) determine the value of \(x\) for which the kinetic encrgy of the system is a maximum, and (c) obtain the child's maximum upward speed.

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