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?

1 Answer
May 25, 2016

Answer:

#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#