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

1 Answer
Aug 12, 2017

Answer:

The work is #=513.1J#

Explanation:

We need

#inttanxdx=ln|cos(x)|+C#

The work done is

#W=F*d#

The frictional force is

#F_r=mu_k*N#

The normal force is #N=mg#

The mass is #m=4kg#

#F_r=mu_k*mg#

#=4(5+tanx)g#

The work done is

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

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

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

#=4g(50/12pi+ln|cos(5/12pi)|-ln|cos(5/12pi)|)#

#=4g(25/6pi+0)#

#=513.1J#