Question

An object with a mass of `4 kg` is pushed along a linear path with a kinetic friction coefficient of `u_k(x)= 5+tanx`. How much work would it take to move the object over `x in [(pi)/12, (pi)/4]`, where x is in meters?

Answer 1

`W~~115"J"`

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 `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`

`=>color(darkblue)(W=int_(x_i)^(x_f)mu_kmgdx)`

We have the following information:

• `|->"m"=4"kg"`
• `|->mu_k(x)=5+tan(x)`
• `|->x in[pi/12,pi/4]`
• `|->g=9.81"m"//"s"^2`

Returning to our integration, know the `mg` quantity, which we can treat as a constant and move outside the integral.

`color(darkblue)(W=mgint_(x_i)^(x_f)mu_kdx)`

Substituting in our known values:

`=>W=(4)(9.81)int_(pi/12)^(pi/4)5+tan(x)dx`

Evaluating:

`=>(4)(9.81)[5x+lnabs(sec(x))]_(pi/12)^(pi/4)`

`=>(39.24)[((5pi)/4+lnabs(sqrt2))-((5pi)/12+lnabs(sec(pi/12))]`

`=>(39.24)[(5pi)/6+lnabs(sqrt2)-lnabs(sec(pi/12)))]`

`=>~~114.93`

Therefore, we have that the work done is `~~115J`.