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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Chapter7: Work, Energy, And Energy Resources
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Problem 64PE: Integrated Concepts (a) What force must be supplied by an elevator cable to produce an acceleration...
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A new type of force was discovered by physicists with the following expression:
a
Fnew =
+ Be* + 3x4
where alpha & beta are constants, and x is the position. The expression above was obtained from the interaction of a massless Higgs Boson (a type of particle) and
black
hole
Quantum physicists then decides to design and build a machine that is able to move the Higgs Boson from xɔ to x1. How much work should the machine do to achieve this feat?
(For simplicity, consider that no energy is lost in the process)
Solution
To determine the work done we apply the following
W =
dx
Evaluating the above, we get
W =
for the limits from xị to xf
substituting x1 and xɔ as the limits, the work done is expressed as
| + BI
X1 -
W =
( x15
x25 )
Transcribed Image Text:A new type of force was discovered by physicists with the following expression: a Fnew = + Be* + 3x4 where alpha & beta are constants, and x is the position. The expression above was obtained from the interaction of a massless Higgs Boson (a type of particle) and black hole Quantum physicists then decides to design and build a machine that is able to move the Higgs Boson from xɔ to x1. How much work should the machine do to achieve this feat? (For simplicity, consider that no energy is lost in the process) Solution To determine the work done we apply the following W = dx Evaluating the above, we get W = for the limits from xị to xf substituting x1 and xɔ as the limits, the work done is expressed as | + BI X1 - W = ( x15 x25 )
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 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 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 K1 results to K2=( K1 Since work done is W=AK=K -K Evaluating results to W=( K1
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