An object with a mass of #6 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 1+5cotx #. How much work would it take to move the object over #x in [(11pi)/12, (7pi)/8]#, where x is in meters?

1 Answer
Aug 17, 2017

Answer:

#W~~107"J"#

Explanation:

Work done by a variable force is defined as the area under the force vs. displacement curve. Therefore, it can be expressed as the integral of force:

#color(darkblue)(W=int_(x_i)^(x_f)F_xdx)#

where #x_i# is the object's initial position and #x_f# is the object's final position

Assuming dynamic equilibrium, where we are applying enough force to move the object but not enough to cause a net force and therefore not enough to cause an acceleration, our parallel forces can be summed as:

#sumF_x=F_a-f_k=0#

Therefore we have that #F_a=f_k#

We also have a state of dynamic equilibrium between our perpendicular forces:

#sumF_y=n-F_g=0#

#=>n=mg#

We know that #vecf_k=mu_kvecn#, so putting it all together, we have:

#vecf_k=mu_kmg#

#=>color(darkblue)(W=int_(x_i)^(x_f)mu_kmgdx)#

We have the following information:

  • #|->"m"=6"kg"#
  • #|->mu_k(x)=1+5cot(x)#
  • #|->x in[(11pi)/12,(7pi)/8]#
  • #|->g=9.81"m"//"s"^2#

Returning to our integration, know the #mg# quantity, which we can treat as a constant and move outside the integral.

#color(darkblue)(W=mgint_(x_i)^(x_f)mu_kdx)#

Substituting in our known values:

#=>W=(6)(9.81)int_((11pi)/12)^((7pi)/8)1+5cot(x)dx#

This is a basic integral.

#=>W~~107.39#

Therefore, we have that the work done is #~~107J#.