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 a black hole. Quantum physicists then decides to design and build a machine that is able to move the Higgs Boson from x2 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 = Fnew dx Evaluating the above, we get W = | + for the limits from x¡ to xf substituting x1 and x2 as the limits, the work done is expressed as W = | + B X1 . ( x15 - x25 )

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Chapter1: Units, Trigonometry. And Vectors
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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 a
black hole.
Quantum physicists then decides to design and build a machine that is able to move the Higgs Boson from x2 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 =
Fnew
dx
x;
Evaluating the above, we get
W =
| +
for the limits from x¡ to Xf
substituting x1 and x2 as the limits, the work done is expressed as
W =
| + B
X1
( 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 a black hole. Quantum physicists then decides to design and build a machine that is able to move the Higgs Boson from x2 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 = Fnew dx x; Evaluating the above, we get W = | + for the limits from x¡ to Xf substituting x1 and x2 as the limits, the work done is expressed as W = | + B X1 ( x15 - x25 )
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