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?

May 25, 2016

$444.84 N m$

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, $F = {F}_{f} = {\mu}_{k} m g = \left(5 + \tan x\right) \times 4 \setminus k g \times 10 \setminus m {s}^{- 2} = \left(200 + 40 \tan x\right) N$

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

${\int}_{- \frac{5 \pi}{12}}^{\frac{\pi}{3}} F \mathrm{dx} = {\int}_{- \frac{5 \pi}{12}}^{\frac{\pi}{3}} \left(200 + 40 \tan x\right) \mathrm{dx} = 200 \left(\frac{\pi}{3} + \frac{5 \pi}{12}\right) - 40 {\left[\ln \left(\cos x\right)\right]}^{\frac{\pi}{3}} {\setminus}_{- \frac{5 \pi}{12}} \approx 471.24 - 40 \times 0.66 = 444.84 N m$

Again, please remember the work done, $W > 444.84 N m$