An object with a mass of $7 k g$ $7 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 1 + x + x cos (\frac{π}{4} - \frac{x}{6})$ $u_k(x)= 1+x+xcos(pi/4-x/6)$ . How much work would it take to move the object over #x in [0, 4pi], where x is in meters?

Question

An object with a mass of $7 k g$ $7 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 1 + x + x cos (\frac{π}{4} - \frac{x}{6})$ $u_k(x)= 1+x+xcos(pi/4-x/6)$ . How much work would it take to move the object over #x in [0, 4pi], where x is in meters?

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Answer 1 · 2017-11-16T07:18:30

The work is $= 10166.5 J$ $=10166.5J$

Explanation:

We need

$\int x cos (\frac{π}{4} - \frac{x}{6}) d x = 18 \sqrt{2} (\frac{x}{6} sin (\frac{x}{6}) - \frac{x}{6} cos (\frac{x}{6}) + cos (\frac{x}{6}) + sin (\frac{x}{6})) + C$ $intxcos(pi/4-x/6)dx=18sqrt2(x/6sin(x/6)-x/6cos(x/6)+cos(x/6)+sin(x/6))+C$

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 $m = 7 k g$ $m=7kg$

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

$= 7 \cdot (1 + x + x cos (\frac{π}{4} - \frac{x}{6})) g$ $=7*(1+x+xcos(pi/4-x/6))g$

The work done is

$W = 7 g \int_{0}^{4 π} (1 + x + x cos (\frac{π}{4} - \frac{x}{6})) d x$ $W=7gint_(0)^(4pi)(1+x+xcos(pi/4-x/6))dx$

$= 7 g \cdot {[x + \frac{x^{2}}{2} + 18 \sqrt{2} (\frac{x}{6} sin (\frac{x}{6}) - \frac{x}{6} cos (\frac{x}{6}) + cos (\frac{x}{6}) + sin (\frac{x}{6}))]}_{0}^{4 π}$ $=7g*[x+x^2/2+18sqrt2(x/6sin(x/6)-x/6cos(x/6)+cos(x/6)+sin(x/6))]_(0)^(4pi)$

$= 7 g (148.2)$ $=7g(148.2)$

$= 10166.5 J$ $=10166.5J$

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An object with a mass of $7 k g$ $7 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 1 + x + x cos (\frac{π}{4} - \frac{x}{6})$ $u_k(x)= 1+x+xcos(pi/4-x/6)$ . How much work would it take to move the object over #x in [0, 4pi], where x is in meters?

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Explanation:

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An object with a mass of 7kg7 kg is pushed along a linear path with a kinetic friction coefficient of uk(x)=1+x+xcos(π4−x6)u_k(x)= 1+x+xcos(pi/4-x/6) . How much work would it take to move the object over #x in [0, 4pi], where x is in meters?

1 Answer

Explanation:

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An object with a mass of $7 k g$ $7 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 1 + x + x cos (\frac{π}{4} - \frac{x}{6})$ $u_k(x)= 1+x+xcos(pi/4-x/6)$ . How much work would it take to move the object over #x in [0, 4pi], where x is in meters?