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 [(-5pi)/12, (pi)/3], where x is in meters?

Answer 1

`444.84 Nm`

Explanation:

Please note the work done is given as, `W = F.s` where `F` denotes force and `s` denotes displacement and both these quantities are vectors.

Also note, the force applied (`F`) in this case must be more than the force of friction (`F_f`). But `F` can be anything more than `F_f`.

So it is not possible to find the work done. What can be found, rather, is the minimum work done in this process.

In that case, `F = F_f = mu_kmg = (5+tanx)xx4\ kgxx10\ ms^{-2} = (200+40tanx) N`

Given, the force is applied on the same direction as the displacement, the work done is given as:

`int_{-{5pi}/12}^{pi/3} Fdx = int_{-{5pi}/12}^{pi/3} (200+40tanx) dx = 200 (pi/3 +{5pi}/12) - 40 [ln(cosx)]^{pi/3}\ _{-{5pi}/12} ~~ 471.24 -40xx0.66 = 444.84 Nm`

Again, please remember the work done, `W>444.84 Nm`