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A moving electron has a Kinetic Energy K1. After a net amount of work is done on it, the electron is moving one-quarter as fast in the opposite direction. What is the work done W in terms of K1? Solution To solve for the work done, first we must determine what is the final kinetic energy of the electron. By concept, we know that K1=(1/2)mv²1 K2=(1/2)mv2 But it was mentioned that: V2=( V1 so, K2 in terms of v1 is K2=( )mv21 Substituting the expression for Kj results to K2=( )K1 Since work done is W=AK=K -K Evaluating results to W=( )K1

A moving electron has a Kinetic Energy K1. After a net amount of work is done on it, the electron is moving one-quarter as fast in the opposite direction. What is the work done W in terms of K1? Solution To solve for the work done, first we must determine what is the final kinetic energy of the electron. By concept, we know that K1=(1/2)mv²1 K2=(1/2)mv2 But it was mentioned that: V2=( V1 so, K2 in terms of v1 is K2=( )mv21 Substituting the expression for Kj results to K2=( )K1 Since work done is W=AK=K -K Evaluating results to W=( )K1

College Physics
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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A moving electron has a Kinetic Energy K1. After a net amount of work is done on it, the electron is moving one-quarter as fast in the opposite direction. What is the work done W in terms of Kq? Solution To solve for the work done, first we must determine what is the final kinetic energy of the electron. By concept, we know that Kq=(1/2)mv²1 K2=(1/2)mv2 But it was mentioned that: V2=( V1 so, K2 in terms of v1 is K2=( )mv²1 Substituting the expression for K1 results to K2=( )K1 Since work done is W=AK=K -K Evaluating results to W=( )K1
Transcribed Image Text:A moving electron has a Kinetic Energy K1. After a net amount of work is done on it, the electron is moving one-quarter as fast in the opposite direction. What is the work done W in terms of Kq? Solution To solve for the work done, first we must determine what is the final kinetic energy of the electron. By concept, we know that Kq=(1/2)mv²1 K2=(1/2)mv2 But it was mentioned that: V2=( V1 so, K2 in terms of v1 is K2=( )mv²1 Substituting the expression for K1 results to K2=( )K1 Since work done is W=AK=K -K Evaluating results to W=( )K1
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