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

1 Answer
Jan 2, 2018

Answer:

The work done is #=588.6J#

Explanation:

#"Reminder : "#

#intsecxdx=ln(tanx+secx)+C#

The work done is

#W=F*d#

The frictional force is

#F_r=mu_k*N#

The coefficient of kinetic friction is #mu_k=(4+secx)#

The normal force is #N=mg#

The mass of the object is #m=7kg#

#F_r=mu_k*mg#

#=7*(4+secx)g#

The work done is

#W=7gint_(-5/12pi)^(1/12pi)(4+secx)dx#

#=7g*[4x+ln(tanx+secx)]_(-5/12pi)^(1/12pi)#

#=7g(1/3pi+ln(tan(pi/12)+sec(pi/12))-(-5/3pi+ln(tan(-5/12pi)+sec(-5/12pi))#

#=7g(8.58)#

#=588.6J#