An object with a mass of #6 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 1+cscx #. How much work would it take to move the object over #x in [(3pi)/4, (7pi)/8], where x is in meters?

1 Answer
May 1, 2018

Answer:

The work is #=66.21J#

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=(1+csc(x))#

The normal force is #N=mg#

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

#F_r=mu_k*mg#

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

The work done is

#W=6gint_(3/4pi)^(7/8pi)(1+csc(x))dx#

#=6g*[x+ln|(tan(x/2))|]_(3/4pi)^(7/8pi)#

#=6g(7/8pi+ln(tan(7/16pi)))-(3/4pi+ln(tan(3/8pi))#

#=6g(1.126)#

#=66.21J#