An object with a mass of #6 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 2+cscx #. How much work would it take to move the object over #x in [pi/2, (3pi)/4], where x is in meters?

1 Answer
Jun 23, 2018

Answer:

The work is #=144.1J#

Explanation:

#"Reminder : "#

#intcscxdx=ln|(tan(x/2))|+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=(2+csc(x))#

The normal force is #N=mg#

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

#F_r=mu_k*mg#

#=6*(2+csc(x))g#

The work done is

#W=6gint_(1/2pi)^(3/4pi)(2+csc(x))dx#

#=6g*[2x+ln|(tan(x/2))|]_(1/2pi)^(3/4pi)#

#=6g(3/2pi+ln(tan(3/8pi)))-(pi+ln(tan(1/4pi))#

#=6g(2.45)#

#=144.1J#