An object with a mass of #3 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 3x^2-2x+12 #. How much work would it take to move the object over #x in [1, 3], where x is in meters?

2 Answers
Aug 5, 2017

Answer:

#W = 1236# #"J"#

Explanation:

We're asked to find the necessary work that needs to be done on a #3#-#"kg"# object to move on the position interval #x in [1color(white)(l)"m", 3color(white)(l)"m"]# with a varying coefficient of kinetic friction #mu_k# represented by the equation

#ul(mu_k(x) = 3x^2-2x+12#

The work #W# done by the necessary force is given by

#ulbar(|stackrel(" ")(" "W = int_(x_1)^(x_2)F_xdx" ")|)# #color(white)(a)# (one dimension)

where

  • #F_x# is the magnitude of the necessary force

  • #x_1# is the original position (#1# #"m"#)

  • #x_2# is the final position (#3# #"m"#)

The necessary force would need to be equal to (or greater than, but we're looking for the minimum value) the retarding friction force, so

#F_x = f_k = mu_kn#

Since the surface is horizontal, #n = mg#, so

#ul(F_x = mu_kmg#

The quantity #mg# equals

#mg = (3color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = 29.43color(white)(l)"N"#

And we also plug in the above coefficient of kinetic friction equation:

#ul(F_x = (29.43color(white)(l)"N")(3x^2-2x+12)#

Work is the integral of force, and we're measuring it from #x=1# #"m"# to #x=3# #"m"#, so we can write

#color(red)(W) = int_(1color(white)(l)"m")^(3color(white)(l)"m")(29.43color(white)(l)"N")(3x^2-2x+12) dx = color(red)(ulbar(|stackrel(" ")(" "1236color(white)(l)"J"" ")|)#

Aug 5, 2017

We observe that Coefficient of kinetic friction is given as

#u_k(x)=3x^2−2x+12#

and that the object is being pushed along a linear path.

We know that force of friction opposes the motion. As such we can infer that movement is only in the #x#-direction.
The angle between force of friction and displacement is #180^@#

Work Done against force #vecF# for moving a distance #vecs# is given as

#W=vecFcdotvecs#

Let the object move a distanace #vecdx#. Work done to move the object against the force of friction

#dW=vec(F_f)cdot vecdx#

Force of friction #vec(F_f)=-u_kNhatx#
where #N# is normal reaction and given #=mg#
#dW=-(3x^2−2x+12)mghatxcdot vecdx#

Integrating both sides for the given interval we get
#W=-int_1^3(3x^2−2x+12)mgdx#

#W=-3xx9.81|3x^3/3−2x^2/2+12x|_1^3#
#W=-29.43[(3^3−3^2+12xx3)-(1^3−1^2+12xx1)]#
#W=-29.43[(27−9+36)-(1−1+12)]#
#W=-29.43(54-12)#
#W=-1236J#

#-ve# sign shows that this amount of energy was lost in doing work against friction.