Question

An object with a mass of `8 kg` is pushed along a linear path with a kinetic friction coefficient of `u_k(x)= x+lnx`. How much work would it take to move the object over `x in [2, 6]`, where `x` is in meters?

Answer 1

The work is `=1674.8J`

Explanation:

We need to compute

`I=intlnxdx`

Apply the integration by parts

`u=lnx`, `=>`, `u'=1/x`

`v'=1`, `=>`, `v=x`

`I=uv-intu'vdx=xlnx-int1/x*xdx`

`=xlnx-int1dx=xlnx-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=8kg`

`F_r=mu_k*mg`

`=8*(x+lnx)g`

The work done is

`W=8gint_(2)^(6)(x+lnx)dx`

`=8g*[1/2x^2+xlnx-x]_(2)^(6)`

`=8g(1/2*36+6ln6-6)-(1/2*4+2ln2-2)`

`=8g(12+6ln6-2ln2)`

`=1674.8J`

The value of `g=9.8ms^-2`