Question

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.

Answer 1

Answer:

a) the acceleration of the particle is (  v$_f$² - v$_i$² ) / 2as

b) the integral W = $\frac{1}{2}$m( $_f$² - v$_i$² )

Explanation:

Given the data in the question;

force on particle F = ma

displacement s = x$_f$ - x$_i$

work done on the particle W = Fs = mas

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

$\frac{1}{2}$m(  v$_f$² - v$_i$² ) = mas

$\frac{1}{2}$(  v$_f$² - v$_i$² ) = as

(  v$_f$² - v$_i$² ) = 2as

a = (  v$_f$² - v$_i$² ) / 2as

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

b) Evaluate the integral W = $\int\limits^{v_{f} }_{v_{i} } mvdv$

$W = \int\limits^{v_{f} }_{v_{i} } mvdv$

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

W = $\frac{1}{2}$m( $_f$² - v$_i$² )

Therefore, the integral W = $\frac{1}{2}$m( $_f$² - v$_i$² )