An object with a mass of $8 k g$ $8 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 2 x^{2} - x + 7$ $u_k(x)= 2x^2-x+7$ . How much work would it take to move the object over #x in [2, 3], where x is in meters?

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Answer 1 · 2017-07-11T14:35:17

The work is $= 1009.4 J$ $=1009.4J$

We need

$\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C (n \neq - 1)$ $intx^ndx=x^(n+1)/(n+1)+C(n!=-1)$

The work done is

$W = F \cdot d$ $W=F*d$

The frictional force is

$F_{r} = μ_{k} \cdot N$ $F_r=mu_k*N$

The normal force is $N = m g$ $N=mg$

The mass is $m = 8 k g$ $m=8kg$

$F_{r} = μ_{k} \cdot m g$ $F_r=mu_k*mg$

$= 8 (2 x^{2} - x + 7) g$ $=8(2x^2-x+7)g$

The work done is

$W = 8 g \int_{2}^{3} (2 x^{2} - x + 7) d x$ $W=8gint_(2)^(3)(2x^2-x+7)dx$

$= 8 g \cdot {[\frac{2}{3} x^{3} - \frac{1}{2} x^{2} + 7 x]}_{2}^{3}$ $=8g*[2/3x^3-1/2x^2+7x]_(2)^(3)$

$= 8 g ((18 - \frac{9}{2} + 21) - (\frac{16}{3} - 2 + 14))$ $=8g((18-9/2+21)-(16/3-2+14))$

$= 8 g (27 - \frac{59}{6})$ $=8g(27-59/6)$

$= 6 g (\frac{103}{6})$ $=6g(103/6)$

$= 1009.4 J$ $=1009.4J$

An object with a mass of 8kg8 kg is pushed along a linear path with a kinetic friction coefficient of uk(x)=2x2−x+7u_k(x)= 2x^2-x+7 . How much work would it take to move the object over #x in [2, 3], where x is in meters?