An object with a mass of #3 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 3x^22x+12 #. How much work would it take to move the object over #x in [1, 3], where x is in meters?
2 Answers
Answer:
Explanation:
We're asked to find the necessary work that needs to be done on a
#ul(mu_k(x) = 3x^22x+12#
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 (#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,
#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")(3x^22x+12)#
Work is the integral of force, and we're measuring it from
#color(red)(W) = int_(1color(white)(l)"m")^(3color(white)(l)"m")(29.43color(white)(l)"N")(3x^22x+12) dx = color(red)(ulbar(stackrel(" ")(" "1236color(white)(l)"J"" "))#
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
The angle between force of friction and displacement is
Work Done against force
#W=vecFcdotvecs#
Let the object move a distanace
#dW=vec(F_f)cdot vecdx#
Force of friction
where
Integrating both sides for the given interval we get