# An object with a mass of 7 kg is pushed along a linear path with a kinetic friction coefficient of mu_k(x)= 4+secx. How much work would it take to move the object over x in [(-5pi)/12, (pi)/4], where x is in meters?

Jul 21, 2017

${w}_{\text{push}} = \underline{7 \left(\frac{8 \pi}{3} \vec{g} + \frac{2 \pi}{3} \vec{g} \sec x\right)}$ $\text{J}$

$x \in \left[- \frac{5 \pi}{12} , \frac{\pi}{4}\right]$

$\vec{g} > 0$

And the work itself is done from the perspective of the worker.

Well, I would use a conservation of energy approach, assuming the object has no kinetic energy (i.e. it does not keep moving if we stop pushing). We just need to overcome the barrier of kinetic friction over a distance $\Delta x$.

$\Delta E = 0 = {w}_{{\vec{F}}_{k}} - {w}_{\text{push}}$

where:

• ${w}_{\text{push}}$ is the work one would have to do to push the object $\frac{\pi}{4} + \frac{5 \pi}{12} = \frac{2 \pi}{3}$ meters forward.
• ${w}_{{\vec{F}}_{k}} = {\vec{F}}_{k} \Delta x$ is the counteracting work done by the kinetic friction.
• ${\vec{F}}_{k} = {\mu}_{k} {\vec{F}}_{N}$ is the friction force, and ${\vec{F}}_{N}$ is the normal force. ${\mu}_{k}$ is the coefficient of kinetic friction.

Here, we do the work from the perspective of ourselves, so ${w}_{\text{push}} < 0$ (we exert energy to do the work), but we incorporated the minus sign into the equation.

As for $\mu \left(x\right)$, we don't have to worry about domain issues, because $\sec x$ is entirely continuous from start to finish in $\left[- \frac{5 \pi}{12} , \frac{\pi}{4}\right]$. $\mu \left(x\right)$ changes as follows:

Including the horizontal sum of the forces:

${\sum}_{i} {\vec{F}}_{x , i} = {\vec{F}}_{\text{push}} - {\vec{F}}_{k} = m {\vec{a}}_{x}$

We realize that by not knowing the acceleration, we couldn't solve for ${\vec{F}}_{\text{push}}$ and obtain ${w}_{\text{push}}$ in this manner.

We do, however, find a use in including the vertical sum of the forces:

${\sum}_{i} {\vec{F}}_{y , i} = {\vec{F}}_{N} - {\vec{F}}_{g} = {\vec{F}}_{N} - m \vec{g} = 0$,

where our convention is that $\vec{g} > 0$ (and the negative is in the subtraction sign), we have:

We now have an expression for ${\vec{F}}_{k}$ in terms of the mass and a constant.

$\textcolor{b l u e}{{w}_{\text{push}}} = {\vec{F}}_{k} \Delta x$

$= {\mu}_{k} {\vec{F}}_{N} \Delta x$

$= \left(4 + \sec x\right) \left(m \vec{g}\right) \left(\frac{2 \pi}{3}\right)$

$= \left(4 + \sec x\right) \left(\text{7 kg" xx g" m/s"^2)((2pi)/3 "m}\right)$

$= \textcolor{b l u e}{7 \left(\frac{8 \pi}{3} g + \frac{2 \pi}{3} g \sec x\right)}$ $\textcolor{b l u e}{\text{J}}$

The work is seen to be a function of the position (but only in $\left[- \frac{5 \pi}{12} , \frac{\pi}{4}\right]$ for its domain).

The initial push should then be hardest, the middle of the distance should be easiest, and then near the end of the distance should be slightly harder.