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)/12, (5pi)/8], where x is in meters?

1 Answer
Jul 24, 2018

Answer:

The work is #=174.64J#

Explanation:

#"Reminder : "#

#intcotxdx=ln|(sin(x))|+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+cot(x))#

The normal force is #N=mg#

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

#F_r=mu_k*mg#

#=6*(1+cot(x))g#

The work done is

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

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

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

#=6g(2.97)#

#=174.64J#