Question

An object with a mass of `4 kg` is pushed along a linear path with a kinetic friction coefficient of `u_k(x)= 5+tanx`. How much work would it take to move the object over #x in [(-pi)/12, (5pi)/12], where x is in meters?

Answer 1

The work is `=359.5J`

Explanation:

`"Reminder : "`

`inttanxdx=ln|(cos(x))|+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=(5+tan(x))`

The normal force is `N=mg`

The mass of the object is `m=4kg`

The acceleration due to gravity is `g=9.8ms^-2`

`F_r=mu_k*mg`

`=4*(5+tan(x))g`

The work done is

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

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

`=4g(25/12pi+ln(cos(5/12pi)))-(-5/12pi+ln(cos(-1/12pi))`

`=4g(9.17)`

`=359.5J`