# 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