An object with a mass of 7 kg7kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 4+secx uk(x)=4+secx. How much work would it take to move the object over #x in [(pi)/8, (pi)/6], where x is in meters?

1 Answer
Sep 6, 2017

The work is =45.93J=45.93J

Explanation:

We need

intsecxdx=ln(tanx+secx)+Csecxdx=ln(tanx+secx)+C

The work done is

W=F*dW=Fd

The frictional force is

F_r=mu_k*NFr=μkN

The normal force is N=mgN=mg

The mass is m=7kgm=7kg

F_r=mu_k*mgFr=μkmg

=7*(4+secx)g=7(4+secx)g

The work done is

W=7gint_(1/8pi)^(1/6pi)(4+secx)dxW=7g16π18π(4+secx)dx

=7g*[4x+ln(tanx+secx)]_(1/8pi)^(1/6pi)=7g[4x+ln(tanx+secx)]16π18π

=7g((2/3pi+ln(tan(pi/6)+sec(pi/6))-(1/2pi+ln(tan(pi/8)+sec(pi/8)))=7g((23π+ln(tan(π6)+sec(π6))(12π+ln(tan(π8)+sec(π8)))

=7g(1/6pi+0.549-0.403)=7g(16π+0.5490.403)

=45.93J=45.93J