An object with a mass of $1 k g$ $1 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 3 x^{2} - x + 2$ $u_k(x)= 3x^2-x+2$ . How much work would it take to move the object over #x in [1, 3], where x is in meters?

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Answer 1 · 2017-08-21T09:14:25

The work is $= 254.8 J$ $=254.8J$

We need

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

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 $m = 1 k g$ $m=1kg$

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

$= 1 \cdot (3 x^{2} - x + 2) g$ $=1*(3x^2-x+2)g$

The work done is

$W = g \int_{1}^{3} (3 x^{2} - x + 2) d x$ $W=gint_(1)^(3)(3x^2-x+2)dx$

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

$= g ((3^{3} - \frac{9}{2} + 6) - (1 - \frac{1}{2} + 2))$ $=g((3^3-9/2+6)-(1-1/2+2))$

$= g (33 - 3 - 4)$ $=g(33-3-4)$

$= (26 g) J$ $=(26g)J$

$= 254.8 J$ $=254.8J$

An object with a mass of 1kg1 kg is pushed along a linear path with a kinetic friction coefficient of uk(x)=3x2−x+2u_k(x)= 3x^2-x+2 . How much work would it take to move the object over #x in [1, 3], where x is in meters?