An object with a mass of #5 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 1-cos(x/2) #. How much work would it take to move the object over #x in [0, 8pi], where x is in meters?

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Mar 9, 2018

Answer:

The work is #=1231.4J#

Explanation:

#"Reminder : "#

#intcosxdx=sinx+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-cos(x/2))#

The normal force is #N=mg#

The mass of the object is #m=5kg#

#F_r=mu_k*mg#

#=5*(1-cos(x/2))g#

The work done is

#W=5gint_(0)^(8pi)(1-cos(x/2))dx#

#=5g*[x-2sin(x/2)]_(0)^(8pi)#

#=5g((8pi-2*0)-(0-2*0))#

#=5g(8pi)#

#=1231.4J#

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