An object with a mass of 1 kg1kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 2e^x-x+3 uk(x)=2exx+3. How much work would it take to move the object over x in [1, 4]x[1,4], where x is in meters?

1 Answer
Aug 14, 2017

W = 1033W=1033 "J"J

Explanation:

We're asked to find the work necessary to push an object on a certain position interval with a varying coefficient of kinetic friction.

The equation for work with a varying force is

W = int_(x_1)^(x_2) F_xcolor(white)(l)dxW=x2x1Fxldx

where

  • x_1x1 and x_2x2 are the initial and final positions

  • F_xFx is the force, which will be equal to the friction force acting (but in the opposite direction):

F_x = mu_kn = mu_kmgFx=μkn=μkmg

So

W = int_(x_1)^(x_2) mu_kmgcolor(white)(l)dxW=x2x1μkmgldx

We know:

  • x_1 = 1x1=1 "m"m

  • x_2 = 4x2=4 "m"m

  • mu_k = 2e^x - x + 3μk=2exx+3

  • m = 1m=1 "kg"kg

  • g = 9.81g=9.81 "m/s"^2m/s2

Plugging these in:

W = int_(1color(white)(l)"m")^(4color(white)(l)"m")(1color(white)(l)"kg")(9.81color(white)(l)"m/s"^2)(2e^x-x+3)color(white)(l)dx = color(red)(ulbar(|stackrel(" ")(" "1033color(white)(l)"J"" ")|)