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# An object with a mass of 5 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 1-cos(x/2) . How much work would it take to move the object over #x in [0, 8pi], where x is in meters?

Mar 9, 2018

#### Answer:

The work is $= 1231.4 J$

#### Explanation:

$\text{Reminder : }$

$\int \cos x \mathrm{dx} = \sin x + C$

The work done is

$W = F \cdot d$

The frictional force is

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

The coefficient of kinetic friction is ${\mu}_{k} = \left(1 - \cos \left(\frac{x}{2}\right)\right)$

The normal force is $N = m g$

The mass of the object is $m = 5 k g$

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

$= 5 \cdot \left(1 - \cos \left(\frac{x}{2}\right)\right) g$

The work done is

$W = 5 g {\int}_{0}^{8 \pi} \left(1 - \cos \left(\frac{x}{2}\right)\right) \mathrm{dx}$

$= 5 g \cdot {\left[x - 2 \sin \left(\frac{x}{2}\right)\right]}_{0}^{8 \pi}$

$= 5 g \left(\left(8 \pi - 2 \cdot 0\right) - \left(0 - 2 \cdot 0\right)\right)$

$= 5 g \left(8 \pi\right)$

$= 1231.4 J$