Question

An object with a mass of `7 kg` is pushed along a linear path with a kinetic friction coefficient of `u_k(x)= 1+x-xcos(x)`. How much work would it take to move the object over #x in [0, 4pi], where x is in meters?

Answer 1

`= 7g ( 4 pi (1+ 2 pi ))`

Explanation:

the simple idea here is that Work = Force `times` Distance [or more specifically `W = int_C \ vec F * d vec x`]

The work is being done against friction, and here, we have for the friction force....

`F_f = mu R = mu m g = 7g mu`

where `mu(x) = 1+x-xcos x`

so

`F_{f} (x) = 7g (1+x-xcos x)`

The work is being done against friction along a linear path ; and thus...

`W = 7g \ int_{x = 0}^{4 pi} dx qquad \ 1+x- color{red}{xcos x}`

The red term can be done by IBP, ie

`int u v' = uv - int u' v`

We have

`u = x, u' = 1`
`v' = cos x, v = sin x`

so

`int dx qquad x cos x = x sin x - int dx quad sin x`
`= x sin x + cos x`

So

`W = 7g \ int_{x = 0}^{4 pi} dx qquad \ 1+x- xcos x`

`= 7g [x+x^2/2- x sin x - cos x]_{x = 0}^{4 pi}`

`= 7g { [4 pi+(4 pi)^2/2 - 1] - [ -1]}`

`= 7g ( 4 pi (1+(4 pi)/2 ))`

`= 7g ( 4 pi (1+ 2 pi )) \ J`