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To understand the meaning and possible applications of the work-energy theorem. In this problem, you will use your prior knowledge to derive one of the most important relationships in mechanics: the work-energy theorem. We will start with a special case: a particle of mass m moving in the x direction at constant acceleration a. During a certain interval of time, the particle accelerates from vi to vf, undergoing displacement s given by s=xf−xi.
A. Find the acceleration a of the particle.
B. Evaluate the integral W = integarvf,vi mudu.

Respuesta :

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Answer:

a) the acceleration of the particle is (  v[tex]_f[/tex]² - v[tex]_i[/tex]² ) / 2as

b) the integral W = [tex]\frac{1}{2}[/tex]m( [tex]_f[/tex]² - v[tex]_i[/tex]² )

Explanation:

Given the data in the question;

force on particle F = ma

displacement s = x[tex]_f[/tex] - x[tex]_i[/tex]

work done on the particle W = Fs = mas

we know that; change in energy = work done       { work energy theorem }

[tex]\frac{1}{2}[/tex]m(  v[tex]_f[/tex]² - v[tex]_i[/tex]² ) = mas

[tex]\frac{1}{2}[/tex](  v[tex]_f[/tex]² - v[tex]_i[/tex]² ) = as

(  v[tex]_f[/tex]² - v[tex]_i[/tex]² ) = 2as

a = (  v[tex]_f[/tex]² - v[tex]_i[/tex]² ) / 2as

Therefore, the acceleration of the particle is (  v[tex]_f[/tex]² - v[tex]_i[/tex]² ) / 2as

b) Evaluate the integral W = [tex]\int\limits^{v_{f} }_{v_{i} } mvdv[/tex]

[tex]W = \int\limits^{v_{f} }_{v_{i} } mvdv[/tex]

[tex]W =m[\frac{v^{2} }{2} ]^{vf}_{vi}[/tex]

W = [tex]\frac{1}{2}[/tex]m( [tex]_f[/tex]² - v[tex]_i[/tex]² )

Therefore, the integral W = [tex]\frac{1}{2}[/tex]m( [tex]_f[/tex]² - v[tex]_i[/tex]² )

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