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

Answer 1

The work is `=513.1J`

Explanation:

We need

`inttanxdx=ln|cos(x)|+C`

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=4kg`

`F_r=mu_k*mg`

`=4(5+tanx)g`

The work done is

`W=4gint_(-5/12pi)^(5/12pi)(5+tanx)dx`

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

`=4g((5*5/12pi+ln|cos(5/12pi)|)-(5*-5/12pi+ln|cos(-5/12pi)|))`

`=4g(50/12pi+ln|cos(5/12pi)|-ln|cos(5/12pi)|)`

`=4g(25/6pi+0)`

`=513.1J`