An object with a mass of #8 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= xlnx #. How much work would it take to move the object over #x in [2, 3], where x is in meters?

1 Answer
Oct 9, 2017

Answer:

The work is #=180.9J#

Explanation:

We need to compute

#I=intxlnxdx#

Apply the integration by parts

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

#v'=x#, #=>#, #v=x^2/2#

#I=uv-intu'vdx=1/2x^2lnx-int1/x*x^2/2dx#

#=1/2x^2lnx-1/2intxdx=1/2x^2lnx-1/2*x^2/2#

#=1/2x^2lnx-1/4x^2+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*(xlnx)g#

The work done is

#W=8gint_(2)^(3)(xlnx)dx#

#=8g*[1/2x^2lnx-1/4x^2]_(2)^(3)#

#=8g(1/2*9*ln3-1/4*9)-(1/2*4ln2-1/4*4)#

#=8g(9/2ln3-2ln2-5/4)#

#=180.9J#