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

1 Answer
Jun 25, 2018

Answer:

The work is #=359.5J#

Explanation:

#"Reminder : "#

#inttanxdx=ln|(cos(x))|+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=(5+tan(x))#

The normal force is #N=mg#

The mass of the object is #m=4kg#

The acceleration due to gravity is #g=9.8ms^-2#

#F_r=mu_k*mg#

#=4*(5+tan(x))g#

The work done is

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

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

#=4g(25/12pi+ln(cos(5/12pi)))-(-5/12pi+ln(cos(-1/12pi))#

#=4g(9.17)#

#=359.5J#