An object with a mass of #6 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 2+cscx #. How much work would it take to move the object over #x in [pi/2, (7pi)/8], where x is in meters?

1 Answer
Jun 15, 2017

Answer:

The work is #=233.5J#

Explanation:

We need

#intcscxdx=ln|tan(x/2)|#

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=6kg#

#F_r=mu_k*mg#

#=6(2+cscx)g#

The work done is

#W=6gint_(1/2pi)^(7/8pi)(2+cscx)dx#

#=6g*[2x+ln|tan(x/2)|]_(1/2pi)^(7/8pi)#

#=6g((2*7/8pi+ln|tan(7/16pi)|)-(2*1/2pi+ln|tan(1/4pi)|))#

#=6g(7/4pi-pi+1.61)#

#=6g(3/4pi+1.61)#

#=233.5J#