An object with a mass of #5 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 1+sin(x/3) #. How much work would it take to move the object over #x in [0, 8pi], where x is in meters?

1 Answer
Jun 14, 2017

Answer:

The work is #=1452J#

Explanation:

The work done is

#W=F*d#

The frictional force is

#F_r=mu_k*N#

The normal force is #N=mg#

The mass is #m=5kg#

#F_r=mu_k*mg#

#=5(1+sin(x/3))g#

The work done is

#W=5gint_(0)^(8pi)(1+sin(x/3))dx#

#=5g*[x-3cos(x/3)]_(0)^(8pi)#

#=5g((8pi+3/2)-(0-3))#

#=5g(8pi+9/2)#

#=5g(29.63)#

#=1452J#