An object with a mass of #8 kg# is pushed along a linear path with a kinetic friction coefficient of #u_k(x)= 5x^2-x+1 #. How much work would it take to move the object over #x in [2, 3], 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:
#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 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
#=>W=int_(x_i)^(x_f)mu_kmgdx#
We know the
#=>W=(8)(9.81)int_(2)^(3)(5x^2-x+1)dx#
This is a basic integral, yielding
Therefore, we have that the work done is