An object with a mass of #5 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= x+xsinx #. How much work would it take to move the object over #x in [0, 8pi]#, where x is in meters?

1 Answer
Aug 10, 2017

Answer:

#W~~14259J#

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"=5"kg"#
  • #|->mu_k(x)=x+xsin(x)#
  • #|->x in[0,8pi]#
  • #|->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=(5)(9.81)int_(0)^(8pi)(x+xsinx)dx#

Let's break the integral up using the sum rule.

#=>=(5)(9.81)[int_(0)^(8pi)(x)dx+int_(0)^(8pi)(xsinx)dx]#

The left is a basic integral, but the right will require integration by parts to solve.

#" "u=x " " du=dx#

#dv=sin(x)dx " " v=-cos(x)#

We will have an integral of the form:

#uv-vintdu#

Therefore, for the right most integral, we obtain:

#-xcos(x)-(int_(0)^(8pi)(-cos x)dx)#

#=>-xcos(x)+int_(0)^(8pi)(cosx)dx#

We now have a basic integral.

Putting it all together:

#(5)(9.81)[int_(0)^(8pi)(x)dx+{-xcos(x)+int_(0)^(8pi)(cosx)dx}]#

Which we can now evaluate, yielding #W=14258.57#

Therefore, we have that the work done is #~~14259J# or #1.4*10^4J#