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^2-xcos(x) . How much work would it take to move the object over x in [pi, 4pi], where x is in meters?

Feb 22, 2018

The work done is $= 45.18 k J$

Explanation:

$\text{Reminder : }$

Integraton by parts

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

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

$v ' = \cos x$, $\implies$, $v = \sin x$

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

The work done is

$W = F \cdot d$

The frictional force is

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

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

The normal force is $N = m g$

The mass of the object is $m = 7 k g$

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

$= 7 \cdot \left(1 + {x}^{2} - x \cos \left(x\right)\right) g$

The work done is

$W = 7 g {\int}_{\pi}^{4 \pi} \left(1 + {x}^{2} - x \cos \left(x\right)\right) \mathrm{dx}$

$= 7 g \cdot {\left[x + \frac{1}{3} {x}^{3} - x \sin \left(x\right) - \cos x\right]}_{\pi}^{4 \pi}$

=7g(4pi+64/3pi^3-4pisin(4pi)-cos(4pi))-(pi+1/3pi^3+1))#

$= 7 g \left(- 2 + 3 \pi + 21 {\pi}^{3}\right)$

$= 45177.2 J$