An object with a mass of 7 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 1+x-xcos(x) . How much work would it take to move the object over #x in [0, 4pi], where x is in meters?

Jun 29, 2016

$= 7 g \left(4 \pi \left(1 + 2 \pi\right)\right)$

Explanation:

the simple idea here is that Work = Force $\times$ Distance [or more specifically $W = {\int}_{C} \setminus \vec{F} \cdot d \vec{x}$]

The work is being done against friction, and here, we have for the friction force....

${F}_{f} = \mu R = \mu m g = 7 g \mu$

where $\mu \left(x\right) = 1 + x - x \cos x$

so

${F}_{f} \left(x\right) = 7 g \left(1 + x - x \cos x\right)$

The work is being done against friction along a linear path ; and thus...

$W = 7 g \setminus {\int}_{x = 0}^{4 \pi} \mathrm{dx} q \quad \setminus 1 + x - \textcolor{red}{x \cos x}$

The red term can be done by IBP, ie

$\int u v ' = u v - \int u ' v$

We have

$u = x , u ' = 1$
$v ' = \cos x , v = \sin x$

so

$\int \mathrm{dx} q \quad x \cos x = x \sin x - \int \mathrm{dx} \quad \sin x$
$= x \sin x + \cos x$

So

$W = 7 g \setminus {\int}_{x = 0}^{4 \pi} \mathrm{dx} q \quad \setminus 1 + x - x \cos x$

$= 7 g {\left[x + {x}^{2} / 2 - x \sin x - \cos x\right]}_{x = 0}^{4 \pi}$

$= 7 g \left\{\left[4 \pi + {\left(4 \pi\right)}^{2} / 2 - 1\right] - \left[- 1\right]\right\}$

$= 7 g \left(4 \pi \left(1 + \frac{4 \pi}{2}\right)\right)$

$= 7 g \left(4 \pi \left(1 + 2 \pi\right)\right) \setminus J$