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 [(-5pi)/12, (pi)/12], where x is in meters?

Answer 1

The work done is `=588.6J`

Explanation:

`"Reminder : "`

`intsecxdx=ln(tanx+secx)+C`

The work done is

`W=F*d`

The frictional force is

`F_r=mu_k*N`

The coefficient of kinetic friction is `mu_k=(4+secx)`

The normal force is `N=mg`

The mass of the object is `m=7kg`

`F_r=mu_k*mg`

`=7*(4+secx)g`

The work done is

`W=7gint_(-5/12pi)^(1/12pi)(4+secx)dx`

`=7g*[4x+ln(tanx+secx)]_(-5/12pi)^(1/12pi)`

`=7g(1/3pi+ln(tan(pi/12)+sec(pi/12))-(-5/3pi+ln(tan(-5/12pi)+sec(-5/12pi))`

`=7g(8.58)`

`=588.6J`