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

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

b) the integral W =
`(1)/(2)`m(
`_f`² - v
`_i`² )

Step-by-step 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 }

`(1)/(2)`m( v
`_f`² - v
`_i`² ) = mas

`(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[(v^(2) )/(2) ]^(vf)_(vi)`

W =
`(1)/(2)`m(
`_f`² - v
`_i`² )

Therefore, the integral W =
`(1)/(2)`m(
`_f`² - v
`_i`² )