Problem A moving electron has a Kinetic Energy Kg. 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 Kg? 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)mv², K2-(1/2)mv2 But it was mentioned that: V2- so, K2 in terms of vq is K2= )mv², Substituting the expression for K1 results to K2- Since work done is w-AK-K -K Evaluating results to W-(
Problem A moving electron has a Kinetic Energy Kg. 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 Kg? 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)mv², K2-(1/2)mv2 But it was mentioned that: V2- so, K2 in terms of vq is K2= )mv², Substituting the expression for K1 results to K2- Since work done is w-AK-K -K Evaluating results to W-(
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Transcribed Image Text:Problem
A moving electron has a Kinetic Energy K7. 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
K1=(1/2)mv²,
K2=(1/2)mv2
But it was mentioned that:
V2=(
so, K2 in terms of v is
K2=(
)mv²1
Substituting the expression for K, results to
K2=(
Since work done is
W=AK=K
-K
Evaluating results to
W=(
)K1
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