An object with a mass of #7 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 1+x+xcos(pi/4-x/6) #. How much work would it take to move the object over #x in [0, 4pi], where x is in meters?

1 Answer
Nov 16, 2017

Answer:

The work is #=10166.5J#

Explanation:

We need

#intxcos(pi/4-x/6)dx=18sqrt2(x/6sin(x/6)-x/6cos(x/6)+cos(x/6)+sin(x/6))+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=7kg#

#F_r=mu_k*mg#

#=7*(1+x+xcos(pi/4-x/6))g#

The work done is

#W=7gint_(0)^(4pi)(1+x+xcos(pi/4-x/6))dx#

#=7g*[x+x^2/2+18sqrt2(x/6sin(x/6)-x/6cos(x/6)+cos(x/6)+sin(x/6))]_(0)^(4pi)#

#=7g(148.2)#

#=10166.5J#