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

1 Answer
Mar 14, 2018

Answer:

The work is #=372.8J#

Explanation:

#"Reminder : "#

#inttanxdx=-ln(cosx)+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+tanx)#

The normal force is #N=mg#

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

#F_r=mu_k*mg#

#=4*(5+tanx)g#

The work done is

#W=4gint_(-3/12pi)^(1/3pi)(5+tanx)dx#

#=4g*[5x-ln(cos(x)]_(-3/12pi)^(1/3pi)#

#=4g((5/3pi-ln(cos(1/3pi))-(-15/12pi-ln(cos(-3/12pi))))#

#=4g(35/12pi+1/12ln64)#

#=4*9.8*9.51#

#=372.8J#