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)/6], where x is in meters?

1 Answer
Aug 31, 2017

Answer:

The work is #=19.5J#

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_(1/12pi)^(1/6pi)(4+secx)dx#

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

#=7g(ln(tan(pi/6)+sec(pi/6))-ln(tan(pi/12)+sec(pi/12))#

#=7g(0.284)#

#=19.5J#