An object with a mass of #6 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 1+5cotx #. How much work would it take to move the object over #x in [(pi)/12, (3pi)/8]#, where #x# is in meters?

1 Answer
Apr 3, 2018

Answer:

The work is #=428.1J#

Explanation:

#"Reminder : "#

#intcotxdx=ln(|sinx|)+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=(1+5cotx)#

The normal force is #N=mg#

The mass of the object is #m=6kg#

#F_r=mu_k*mg#

#=6*(1+5cotx)g#

The work done is

#W=6gint_(1/12pi)^(3/8pi)(1+5cot)dx#

#=6g*[x+5ln(sin(x) ] _(1/12pi)^(3/8pi)#

#=6g((3/8pi+5ln(sin(3/8pi))-(1/12pi+5ln(sin(1/12pi))#

#=6xx9.8xx7.28#

#=428.1J#