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

1 Answer
Feb 25, 2018

Answer:

The work is #=25.28J#

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+5cot(x))#

The normal force is #N=mg#

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

#F_r=mu_k*mg#

#=6*(1+5cot(x))g#

The work done is

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

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

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

#=6g(*0.43)#

#=25.28J#