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 [(-3pi)/12, (pi)/3], where x is in meters?

Answer 1

The work is `=372.8J`

Explanation:

`"Reminder : "`

`inttanxdx=-ln(cosx)+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+tanx)`

The normal force is `N=mg`

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

`F_r=mu_k*mg`

`=4*(5+tanx)g`

The work done is

`W=4gint_(-3/12pi)^(1/3pi)(5+tanx)dx`

`=4g*[5x-ln(cos(x)]_(-3/12pi)^(1/3pi)`

`=4g((5/3pi-ln(cos(1/3pi))-(-15/12pi-ln(cos(-3/12pi))))`

`=4g(35/12pi+1/12ln64)`

`=4*9.8*9.51`

`=372.8J`