An object with a mass of #6 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 1+cotx #. How much work would it take to move the object over #x in [(pi)/4, (7pi)/8], where x is in meters?

1 Answer
Jul 21, 2017

Answer:

The work is #=79.4J#

Explanation:

We need

#intcotxdx=ln|sin(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=6kg#

#F_r=mu_k*mg#

#=6(1+cotx)g N#

The work done is

#W=6gint_(1/4pi)^(7/8pi)(1+cotx)dx#

#=6g*[x+ln|sin(x)|]_(1/4pi)^(7/8pi)#

#=6g((7/8pi+ln|sin(7/8pi)|))-(1/4pi+ln|sin(1/4pi)|))#

#=6g(5/8pi-0.614)#

#=6g(1.35)#

#=79.4J#