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

1 Answer
Jun 22, 2017

Answer:

The work is #=180J#

Explanation:

We need

#intsecxdx=ln(tan(x)+sec(x))#

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_(-1/12pi)^(1/12pi)(4+secx)dx#

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

#=7g((4*1/12pi+ln(tan(1/12pi)+sec(1/12pi))-(-4*1/12pi+ln(tan(-1/12pi)+sec(-1/12pi)))#

#=7g(2/3pi+0.265+0.265)#

#=7g(2.624)#

#=180J#