### Physics Problem: Work Done on a Spring#### Problem Statement:2. A spring on a horizontal surface can be stretched and held 0.5 meters from its equilibrium position with a force of 50 N.##### Part (a)a) How much work is done in stretching the spring 1.5 meters from its equilibrium position?##### Part (b)b) How much work is done in compressing the spring 0.5 meters from its equilibrium position?### ExplanationWhen dealing with springs, Hooke's Law is fundamental. According to Hooke's Law, the force required to stretch or compress a spring is directly proportional to the displacement from its equilibrium position:\[ F = kx \]where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position.The work done in stretching or compressing a spring is given by the formula:\[ W = \frac{1}{2} k x^2 \]To solve these problems, we'll first need to determine the spring constant \( k \).### Part (a) SolutionGiven:- Force \( F = 50 \) N- Displacement \( x = 0.5 \) metersUsing Hooke's Law:\[ 50 = k \times 0.5 \]\[ k = \frac{50}{0.5} \]\[ k = 100 \, \text{N/m} \]Next, we need to find the work done in stretching the spring 1.5 meters:\[ W = \frac{1}{2} k x^2 \]\[ W = \frac{1}{2} \times 100 \times (1.5)^2 \]\[ W = 0.5 \times 100 \times 2.25 \]\[ W = 112.5 \, \text{J} \]### Part (b) SolutionTo find the work done in compressing the spring 0.5 meters, again we use:\[ W = \frac{1}{2} k x^2 \]\[ W = \frac{1}{2} \times 100 \times (0.5)^2 \]\[ W = 0.5 \times 100 \times 0.25 \]\[ W = 12.5 \, \text

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Description: ### Physics Problem: Work Done on a Spring#### Problem Statement:2. A spring on a horizontal surface can be stretched and held 0.5 meters from its equilibrium position with a force of 50 N.##### Part (a)a) How much work is done in stretching the spr

Given: The stretching in the spring due to force 50 N is 0.5 m from the equilibrium.

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2. A spring on a horizontal surface can be stretched and held 0.5 meters from its equilibrium position with a force of 50 N. a) How much work is done in stretching the spring 1.5 meters from its equilibrium position? b) How much work is done in compressing the spring 0.5 meters from its equilibrium position?

2. A spring on a horizontal surface can be stretched and held 0.5 meters from its equilibrium position with a force of 50 N. a) How much work is done in stretching the spring 1.5 meters from its equilibrium position? b) How much work is done in compressing the spring 0.5 meters from its equilibrium position?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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