An object with a mass of #3 kg# is pushed along a linear path with a kinetic friction coefficient of #mu_k(x)= x^23x+6 #. How much work would it take to move the object over #x in [2, 3]#, where #x# is in meters?
1 Answer
Explanation:
We're asked to find the necessary work that needs to be done on a
#ul(mu_k(x) = x^23x+6#
The work
#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 (#2# #"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,
#ul(F_x = mu_kmg#
The quantity
#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")(x^23x+6)#
Work is the integral of force, and we're measuring it from
#color(red)(W) = int_(2color(white)(l)"m")^(3color(white)(l)"m")(29.43color(white)(l)"N")(x^23x+6) dx = color(red)(ulbar(stackrel(" ")(" "142color(white)(l)"J"" "))#