An object with a mass of #4 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 1+3cscx #. How much work would it take to move the object over #x in [(pi)/6, (pi)/4], where x is in meters?

Question

1032 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-24T11:27:21

The work is #=61.5J#

#"Reminder : "#

#intcscxdx=ln|(tan(x/2))|+C#

The work done is

#W=F*d#

The frictional force is

#F_r=mu_k*N#

The coefficient of kinetic friction is #mu_k=(1+3csc(x))#

The normal force is #N=mg#

The mass of the object is #m=4kg#

#F_r=mu_k*mg#

#=4*(1+3csc(x))g#

The work done is

#W=4gint_(1/6pi)^(1/4pi)(1+3csc(x))dx#

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

#=4g(1/4pi+ln(tan(1/8pi)))-(1/6pi+ln(tan(1/12pi))#

#=4g(1.57)#

#=61.5J#