An object with a mass of 3 kg3kg is pushed along a linear path with a kinetic friction coefficient of mu_k(x)= x^2-3x+6 μk(x)=x2−3x+6. How much work would it take to move the object over x in [2, 3]x∈[2,3], where xx 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^2-3x+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^2-3x+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^2-3x+6) dx = color(red)(ulbar(|stackrel(" ")(" "142color(white)(l)"J"" ")|)