An object with a mass of #7 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 4+secx #. How much work would it take to move the object over #x in [(-3pi)/12, (pi)/6], where x is in meters?

1 Answer
Dec 25, 2017

Answer:

The work is #=457.6J#

Explanation:

We need

#intsecxdx=ln(tanx+secx)+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=7kg#

#F_r=mu_k*mg#

#=7*(4+secx)g#

The work done is

#W=7gint_(-3/12pi)^(1/6pi)(4+secx)dx#

#=7g*[4x+ln(tanx+secx)]_(-3/12pi)^(1/6pi)#

#=7g(4/6pi+ln(tan(pi/6)+sec(pi/6))-(-pi+ln(tan(-1/4pi)+sec(-1/4pi))#

#=7g(6.67)#

#=457.6J#