An object with a mass of $8 k g$ $8 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 2 x^{2} - x$ $u_k(x)= 2x^2-x$ . 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 · 2017-10-28T06:16:26

The work is $= 797.1 J$ $=797.1J$

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 $m = 8 k g$ $m=8kg$

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

$= 8 \cdot (2 x^{2} - x) g$ $=8*(2x^2-x)g$

The acceleration due to gravity is $g = 9.8 m s^{- 2}$ $g=9.8ms^-2$

The work done is

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

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

$= 8 g (\frac{2}{3} \cdot 3^{3} - \frac{9}{2})) - (\frac{2}{3} \cdot 8 - 2)$ $=8g(2/3*3^3-9/2))-(2/3*8-2)$

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

$= 8 g (20 - \frac{59}{6})$ $=8g(20-59/6)$

$= 797.1 J$ $=797.1J$

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