Question

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?

Answer 1

`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`