An object with a mass of $5 k g$ $5 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = x e^{x} - x$ $u_k(x)= xe^x-x$ . How much work would it take to move the object over #x in [1, 2], where x is in meters?

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Answer 1 · 2017-08-23T05:13:40

The work is $= 288.6 J$ $=288.6J$

We need

$\int x e^{x} d x = x e^{x} - e^{x} + C$ $intxe^xdx=xe^x-e^x+C$

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 $m = 5 k g$ $m=5kg$

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

$= 5 \cdot (x e^{x} - x) g$ $=5*(xe^x-x)g$

The work done is

$W = 5 g \int_{1}^{2} (x e^{x} - x) d x$ $W=5gint_(1)^(2)(xe^x-x)dx$

$= 5 g \cdot {[x e^{x} - e^{x} - \frac{x^{2}}{2}]}_{1}^{x}$ $=5g*[xe^x-e^x-x^2/2]_(1)^(2)$

$= 5 g ((2 e^{2} - e^{2} - 2) - (e - e - \frac{1}{2}))$ $=5g((2e^2-e^2-2)-(e-e-1/2))$

$= 5 g (e^{2} - 2 + \frac{1}{2}) = 5 g (e^{2} - \frac{3}{2})$ $=5g(e^2-2+1/2)=5g(e^2-3/2)$

$= 288.6 J$ $=288.6J$

An object with a mass of 5kg5 kg is pushed along a linear path with a kinetic friction coefficient of uk(x)=xex−xu_k(x)= xe^x-x . How much work would it take to move the object over #x in [1, 2], where x is in meters?