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?

1 Answer
Oct 17, 2017

Answer:

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#