Question

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?

Answer 1

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`