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, (pi)/4]#, where #x# is in meters?

1 Answer
Nov 1, 2017

Answer:

The work is #=217.6J#

Explanation:

We need

#inttanxdx=-ln(|cosx|)+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=4kg#

#F_r=mu_k*mg#

#=4*(5+tanx)g#

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

The work done is

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

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

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

#=4g(5.55))#

#=217.6J#