/*! 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 917 When a \(4 \mathrm{~kg}\) mass i... [FREE SOLUTION] | 91Ó°ÊÓ

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When a \(4 \mathrm{~kg}\) mass is hung vertically on a light spring that obeys Hook's law, the spring stretches by \(2 \mathrm{~cm}\) what will be the work required to be done by an external agent in stretching this spring by \(5 \mathrm{~cm}\) ? (A) \(4.900\) Joule (B) \(2.450\) Joule (C) \(0.495\) Joule (D) \(0.245\) Joule

Short Answer

Expert verified
The work required to be done by an external agent in stretching the spring by 5 cm is approximately 2.0621 Joules, and the closest option is (B) 2.450 Joule.

Step by step solution

01

Identify the Given Information and Knowns

According to the problem, we know the following information: Mass, m = 4 kg Initial spring stretch, x1 = 2 cm (0.02 m, converted from cm to meters) Additional stretch, x2 = 5 cm (0.05 m, converted from cm to meters) From Hooke's law, F = kx, we need to find the spring constant, 'k', and then calculate the work done, W.
02

Calculate the Spring Constant

From Hooke's law, F = kx, and we know the force acting on the spring is the gravitational force due to the mass (m) hung on the spring, which is F = mg. Now, substitute this force and the initial stretch into Hooke's law: mg = kx1 4 kg × 9.81 m/s² = k × 0.02 m Now solve for k: k = (4 kg × 9.81 m/s²) / 0.02 m k = 1962 N/m
03

Calculate the Work Done

To find the work done, we use the formula for work done on a spring, which is W = (1/2)k(x2² - x1²). Plug in the values for k, x1, and x2: W = (1/2)(1962 N/m)(0.05 m² - 0.02 m²) W = (1/2)(1962 N/m)(0.0025 m² - 0.0004 m²) W = (1/2)(1962 N/m)(0.0021 m²) W = 2.0621 J The work required to be done by an external agent in stretching the spring by 5 cm is approximately 2.0621 Joules. The closest answer among the given options is (B) 2.450 Joule.

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Key Concepts

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

Hooke's Law
When studying springs and their behavior, one cannot overlook Hooke's Law. It is a fundamental principle that describes how springs stretch or compress under an applied force. Hooke's Law is usually expressed by the formula:\[ F = kx \]Where:
  • \( F \) is the force applied to the spring (in newtons),
  • \( k \) is the spring constant (in newtons per meter), and
  • \( x \) is the displacement of the spring from its equilibrium position (in meters).
This law tells us that the force needed to extend or compress a spring is directly proportional to the distance the spring is stretched or compressed. The main idea is that springs resist deformation, which means the more you stretch them, the harder they push back.
Spring Constant
The spring constant, represented by \( k \), is a critical parameter in Hooke's Law. It indicates how stiff or flexible a spring is. A higher spring constant means the spring is stiffer, while a lower spring constant indicates a more easily stretchable spring. You can calculate the spring constant using the equation derived from Hooke's Law.If you have a known force and a resulting displacement, you find \( k \) by rearranging the formula:\[ k = \frac{F}{x} \]For example, if a 4 kg mass stretches a spring by 2 cm, the force is the weight of the object (mass times gravity), and so:\[ F = 4 \times 9.81 = 39.24 \text{ N} \]Substituting this force and the displacement into the formula gives:\[ k = \frac{39.24}{0.02} = 1962 \text{ N/m} \]Understanding the spring constant helps in predicting how a spring will behave under different weights or forces.
Elastic Potential Energy
Elastic potential energy is the energy stored in elastic materials as a result of their stretching or compressing. When a spring is stretched or compressed, it stores energy that can be converted back into kinetic energy, or work, when the spring reverts to its resting state. The formula for calculating elastic potential energy in a spring is:\[ E_{spring} = \frac{1}{2}kx^2 \]Where:
  • \( E_{spring} \) is the elastic potential energy,
  • \( k \) is the spring constant, and
  • \( x \) is the displacement from equilibrium.
The formula highlights that the energy stored is directly proportional to the square of the displacement, meaning even small changes in stretch can lead to significant energy storage when the spring constant is large.
Work-Energy Principle
The Work-Energy Principle is a key concept in physics that establishes a connection between work done and energy changes in a system. According to this principle, the work done by all forces acting on a particle equals the change in its kinetic energy. This concept can be applied to understand the work done on a spring. To find the work done in stretching or compressing a spring, you can use the formula:\[ W = \frac{1}{2}k(x_2^2 - x_1^2) \]Here:
  • \( W \) is the work done,
  • \( k \) is the spring constant,
  • \( x_1 \) is the initial displacement, and
  • \( x_2 \) is the final displacement.
By calculating the difference in the potential energy at the two points, this formula helps determine the work done in stretching a spring from one displacement to another. Understanding this principle allows you to predict how energy transfers in systems involving springs.

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