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)/4], where x is in meters?

1 Answer
Oct 30, 2017

Answer:

The work is #=307.9J#

Explanation:

We need

#inttanxdx=-ln(|cosx|)+C #

#-3/12pi=-1/4pi#

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_(-1/4pi)^(1/4pi)(5+tanx)dx#

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

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

#=4g(5/2pi))#

#=10gpi#

#=307.9J#