Question

An object with a mass of `7 kg` is pushed along a linear path with a kinetic friction coefficient of `u_k(x)= 4+secx`. How much work would it take to move the object over #x in [(-pi)/12, (pi)/12], where x is in meters?

Answer 1

The work is `=180J`

Explanation:

We need

`intsecxdx=ln(tan(x)+sec(x))`

The work done is

`W=F*d`

The frictional force is

`F_r=mu_k*N`

The normal force is `N=mg`

The mass is `m=7kg`

`F_r=mu_k*mg`

`=7(4+secx)g`

The work done is

`W=7gint_(-1/12pi)^(1/12pi)(4+secx)dx`

`=7g*[4x+ln(tanx+secx)]_(-1/12pi)^(1/12pi)`

`=7g((4*1/12pi+ln(tan(1/12pi)+sec(1/12pi))-(-4*1/12pi+ln(tan(-1/12pi)+sec(-1/12pi)))`

`=7g(2/3pi+0.265+0.265)`

`=7g(2.624)`

`=180J`