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)/12], where x is in meters?

1 Answer
Jun 22, 2017

The work is =180J=180J

Explanation:

We need

intsecxdx=ln(tan(x)+sec(x))secxdx=ln(tan(x)+sec(x))

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)^(1/12pi)(4+secx)dxW=7g112π112π(4+secx)dx

=7g*[4x+ln(tanx+secx)]_(-1/12pi)^(1/12pi)=7g[4x+ln(tanx+secx)]112π112π

=7g((4*1/12pi+ln(tan(1/12pi)+sec(1/12pi))-(-4*1/12pi+ln(tan(-1/12pi)+sec(-1/12pi)))=7g((4112π+ln(tan(112π)+sec(112π))(4112π+ln(tan(112π)+sec(112π)))

=7g(2/3pi+0.265+0.265)=7g(23π+0.265+0.265)

=7g(2.624)=7g(2.624)

=180J=180J