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 [(3pi)/12, (5pi)/12], where x is in meters?

1 Answer
Jan 20, 2018

Answer:

The work is #=141.9J#

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#

#F_r=mu_k*mg#

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

The work done is

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

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

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

#=4g(3.62)#

#=141.9J#