An object with a mass of #4 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 5+tanx #. How much work would it take to move the object over #x in [(pi)/12, (5pi)/12], where x is in meters?

1 Answer
Jun 6, 2017

Answer:

The work is #=257J#

Explanation:

The work done is

#W=F*d#

The frictional force is

#F_r=mu_k*N#

The normal force is #N=mg#

#F_r=mu_k*mg#

#=4(5+tanx))g#

The work done is

#W=4gint_(1/12pi)^(5/12pi)(5+tanx)dx#

#=4g*[5x-ln|cosx|)]_(1/12pi)^(5/12pi)#

#=4g((25/12pi-ln|cos(5/12pi)|)-(5/12pi-ln|cos(1/12pi)|))#

#=4g(5/3pi-ln|cos(5/12pi)|+ln(cos(1/12pi))#

#=4g(5/3pi+1.32))#

#=4g*6.56J#

#=257J#