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?

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Chapter1: Units, Trigonometry. And Vectors
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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?

Write powers or subscript as is. Ex: Use b2 if you mean
b2
or b2
Write transcendental functions as is. Ex. Use costheta if you mean cose.
Spell out Greek letters. Ex: Use pi if you mean A
Write answers without spaces. Ex: Use 2epsilonOr3 if you mean 2ɛ r³.
Write fractions with a slash. Ex: Use 1/2 if you mean
Transcribed Image Text:Write powers or subscript as is. Ex: Use b2 if you mean b2 or b2 Write transcendental functions as is. Ex. Use costheta if you mean cose. Spell out Greek letters. Ex: Use pi if you mean A Write answers without spaces. Ex: Use 2epsilonOr3 if you mean 2ɛ r³. Write fractions with a slash. Ex: Use 1/2 if you mean
Problem
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=(
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:Problem 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=( 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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