An object with a mass of #6 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 1+5cotx #. How much work would it take to move the object over #x in [(11pi)/12, (5pi)/8], where x is in meters?

1 Answer
Apr 2, 2017

Answer:

The work is #=320.5J#

Explanation:

The mass is #m=6kg#

The nom'rmal reaction is #N=6gN#

The coefficient of kinetic friction is

#mu_k=F_r/N#

#N=mg#

So,

#F_r=mu_k*N#

#=(1+5cotx)*6g#

The work done is

#W=F_r*d#

#=int_(11/12pi)^(5/8pi)(1+5cotx)*6gdx#

#=6gint_(11/12pi)^(5/8pi)(1+5cotx)dx#

The integral of #cotx# is

#=intcotxdx=int(cosxdx)/sinx#

#=ln(|sinx|)#

So,

#W=6g*[x+5ln(|sinx|)]_(11/12pi)^(5/8pi)#

#=6g*((5/8pi+5ln(|sin(5/8pi)|)-(11/12pi+5ln(|sin(11/12pi|)))#

#=6g*(-11/12pi+5/8pi+5(lnsin(5/8pi)-lnsin(11/12pi))#

#=6g*5.45#

#=320.5J#