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)/12, (pi)/6], where x is in meters?

1 Answer
Aug 31, 2017

The work is =19.5J=19.5J

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_(1/12pi)^(1/6pi)(4+secx)dxW=7g16π112π(4+secx)dx

=7g*[ln(tanx+secx)]_(1/12pi)^(1/6pi)=7g[ln(tanx+secx)]16π112π

=7g(ln(tan(pi/6)+sec(pi/6))-ln(tan(pi/12)+sec(pi/12))=7g(ln(tan(π6)+sec(π6))ln(tan(π12)+sec(π12))

=7g(0.284)=7g(0.284)

=19.5J=19.5J