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

1 Answer
Aug 14, 2017

Answer:

#W = 1033# #"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)dx#

where

  • #x_1# and #x_2# are the initial and final positions

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

#F_x = mu_kn = mu_kmg#

So

#W = int_(x_1)^(x_2) mu_kmgcolor(white)(l)dx#

We know:

  • #x_1 = 1# #"m"#

  • #x_2 = 4# #"m"#

  • #mu_k = 2e^x - x + 3#

  • #m = 1# #"kg"#

  • #g = 9.81# #"m/s"^2#

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"" ")|)#