# An object with a mass of 6 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 1+5cotx . How much work would it take to move the object over x in [(11pi)/12, (5pi)/8], where x is in meters?

Apr 2, 2017

The work is $= 320.5 J$

#### Explanation:

The mass is $m = 6 k g$

The nom'rmal reaction is $N = 6 g N$

The coefficient of kinetic friction is

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

$N = m g$

So,

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

$= \left(1 + 5 \cot x\right) \cdot 6 g$

The work done is

$W = {F}_{r} \cdot d$

$= {\int}_{\frac{11}{12} \pi}^{\frac{5}{8} \pi} \left(1 + 5 \cot x\right) \cdot 6 g \mathrm{dx}$

$= 6 g {\int}_{\frac{11}{12} \pi}^{\frac{5}{8} \pi} \left(1 + 5 \cot x\right) \mathrm{dx}$

The integral of $\cot x$ is

$= \int \cot x \mathrm{dx} = \int \frac{\cos x \mathrm{dx}}{\sin} x$

$= \ln \left(| \sin x |\right)$

So,

$W = 6 g \cdot {\left[x + 5 \ln \left(| \sin x |\right)\right]}_{\frac{11}{12} \pi}^{\frac{5}{8} \pi}$

=6g*((5/8pi+5ln(|sin(5/8pi)|)-(11/12pi+5ln(|sin(11/12pi|)))

=6g*(-11/12pi+5/8pi+5(lnsin(5/8pi)-lnsin(11/12pi))#

$= 6 g \cdot 5.45$

$= 320.5 J$