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/12, (3pi)/4], where x is in meters?

Question

1149 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-04T07:53:45

The work is #=417.5Jd#

We need

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

The work done is

#W=F*d#

The frictional force is

#F_r=mu_k*N#

The normal force is #N=mg#

So, #F_r=mu_k*mg#

#=6(2+cscx))g#

The work done is

#W=6gint_(1/12pi)^(3/4pi)(2+cscx)dx#

#=6g*[2x+ln|tan(x/2)|]_(1/12pi)^(3/4pi)#

#=6g((3/2pi+ln|tan(3/8pi)|)-(1/6pi+ln|tan(pi/24)|)#

#=6g(4/3pi+0.88+2.027)#

#=6g(7.098)#

#=42.6gJ#

#=417.5J#