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
Aug 6, 2017

Answer:

#W~~2367J#

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 #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#.

#=>W=int_(x_i)^(x_f)mu_kmgdx#

We know the #mg# quantity, which we can treat as a constant and move outside the integral.

#=>W=(8)(9.81)int_(2)^(3)(5x^2-x+1)dx#

This is a basic integral, yielding #W=2367.48#

Therefore, we have that the work done is #~~2367J# or #2.4*10^3J#