An object with a mass of #5 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= e^x2x+3 #. How much work would it take to move the object over #x in [3, 4], where x is in meters?
1 Answer
Answer:
Explanation:
We're asked to find the necessary work that needs to be done on a
#ul(mu_k(x) = e^x  2x + 3#
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 (#3# #"m"# ) 
#x_2# is the final position (#4# #"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 = (5color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = 49.05color(white)(l)"N"#
And we also plug in the above coefficient of kinetic friction equation:
#ul(F_x = (49.05color(white)(l)"N")(e^x2x+3)#
Work is the integral of force, and we're measuring it from
#color(red)(W) = int_(3color(white)(l)"m")^(4color(white)(l)"m")(49.05color(white)(l)"N")(e^x2x+3) dx = color(red)(ulbar(stackrel(" ")(" "1497color(white)(l)"J"" "))#