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
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
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_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
#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.
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
Therefore, we have that the work done is
