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^2-xcos(x) #. How much work would it take to move the object over #x in [pi, 4pi], where x is in meters?

Question

1286 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-22T13:02:30

The work done is #=45.18kJ#

#"Reminder : "#

Integraton by parts

#intuv'=uv-intu'v#

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

#v'=cosx#, #=>#, #v=sinx#

#intxcosxdx=xsinx+cosx+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+x^2-xcos(x))#

The normal force is #N=mg#

The mass of the object is #m=7kg#

#F_r=mu_k*mg#

#=7*(1+x^2-xcos(x))g#

The work done is

#W=7gint_(pi)^(4pi)(1+x^2-xcos(x))dx#

#=7g*[x+1/3x^3-xsin(x)-cosx]_(pi)^(4pi)#

#=7g(4pi+64/3pi^3-4pisin(4pi)-cos(4pi))-(pi+1/3pi^3+1))#

#=7g(-2+3pi+21pi^3)#

#=45177.2J#