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

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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
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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 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

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