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)/12, (pi)/4], where x is in meters?

1 Answer
Jan 15, 2018

Answer:

The work is #=155.2J#

Explanation:

#"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_(1/12pi)^(1/4pi)(1+3csc(x))dx#

#=4g*[x+3ln|(tan(x/2))|]_(1/12pi)^(1/4pi)#

#=4g(1/4pi+3ln(tan(pi/8)))-(1/12pi+3ln(tan(pi/24))#

#=4g(3.96)#

#=155.2J#