An object with a mass of 6 kg6kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 2+cscx uk(x)=2+cscx. How much work would it take to move the object over #x in [pi/4, (3pi)/4], where x is in meters?

1 Answer
Dec 22, 2017

The work is =288.1J=288.1J

Explanation:

We need

intcscxdx=ln|tan(x/2)|+Ccscxdx=lntan(x2)+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=6kgm=6kg

F_r=mu_k*mgFr=μkmg

=6*(2+cscx)g=6(2+cscx)g

The work done is

W=6gint_(1/4pi)^(3/4pi)(2+cscx)dxW=6g34π14π(2+cscx)dx

=6g*[2x+ln|tan(x/2)|]_(1/4pi)^(3/4pi)=6g[2x+lntan(x2)]34π14π

=6g((3/2pi+ln|tan(3/8pi)|+sec(pi/6))-(1/2pi+ln|tan(pi/8)|)=6g((32π+lntan(38π)+sec(π6))(12π+lntan(π8))

=6g(*4.9)=6g(4.9)

=288.1J=288.1J