An object with a mass of #7 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 1+x-xcos(x/6) #. How much work would it take to move the object over #x in [0, 4pi], where x is in meters?

1 Answer
Nov 25, 2017

Answer:

The work is #=5501.7J#

Explanation:

We need to calculate the integral #intxcos(x/6)dx#

Perform this integration by parts

Let #u=x#, #=>#, #u'=1#

#v'=cos(x/6)#, #=>#, #v=6sin(x/6)#

#intuv'dx=uv-intu'vdx#

#intxcos(x/6)dx=6xsin(x/6)-6intsin(x/6)dx#

#=6xsin(x/6)+36cos(x/6)#

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=7kg#

The acceleration due to gravity is #g=9.8ms^-2#

#F_r=mu_k*mg#

#=7*(1+x-xcos(x/6))g#

The work done is

#W=7gint_(0)^(4pi)(1+x-xcos(x/6))dx#

#=7g[x+1/2x^2-6xsin(x/6)-36cos(x/6)]_0^(4pi)#

#=7g((4pi+1/2(4pi)^2-24pisin(2/3pi)-36cos(2/3pi))-(0+0-0-36))#

#=7g(80.2)#

#=5501.7J#