An object with a mass of 8 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= x+lnx . How much work would it take to move the object over x in [2, 6], where x is in meters?

Oct 17, 2017

The work is $= 1674.8 J$

Explanation:

We need to compute

$I = \int \ln x \mathrm{dx}$

Apply the integration by parts

$u = \ln x$, $\implies$, $u ' = \frac{1}{x}$

$v ' = 1$, $\implies$, $v = x$

$I = u v - \int u ' v \mathrm{dx} = x \ln x - \int \frac{1}{x} \cdot x \mathrm{dx}$

$= x \ln x - \int 1 \mathrm{dx} = x \ln x - x + C$

The work done is

$W = F \cdot d$

The frictional force is

${F}_{r} = {\mu}_{k} \cdot N$

The normal force is $N = m g$

The mass is $m = 8 k g$

${F}_{r} = {\mu}_{k} \cdot m g$

$= 8 \cdot \left(x + \ln x\right) g$

The work done is

$W = 8 g {\int}_{2}^{6} \left(x + \ln x\right) \mathrm{dx}$

$= 8 g \cdot {\left[\frac{1}{2} {x}^{2} + x \ln x - x\right]}_{2}^{6}$

$= 8 g \left(\frac{1}{2} \cdot 36 + 6 \ln 6 - 6\right) - \left(\frac{1}{2} \cdot 4 + 2 \ln 2 - 2\right)$

$= 8 g \left(12 + 6 \ln 6 - 2 \ln 2\right)$

$= 1674.8 J$

The value of $g = 9.8 m {s}^{-} 2$