An object with a mass of #4 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 5+tanx #. How much work would it take to move the object over #x in [(pi)/12, (pi)/4]#, where x is in meters?

1 Answer
Aug 17, 2017

Answer:

#W~~115"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"=4"kg"#
  • #|->mu_k(x)=5+tan(x)#
  • #|->x in[pi/12,pi/4]#
  • #|->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=(4)(9.81)int_(pi/12)^(pi/4)5+tan(x)dx#

Evaluating:

#=>(4)(9.81)[5x+lnabs(sec(x))]_(pi/12)^(pi/4)#

#=>(39.24)[((5pi)/4+lnabs(sqrt2))-((5pi)/12+lnabs(sec(pi/12))]#

#=>(39.24)[(5pi)/6+lnabs(sqrt2)-lnabs(sec(pi/12)))]#

#=>~~114.93#

Therefore, we have that the work done is #~~115J#.