An object with a mass of $6 k g$ $6 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 1 + 5 cot x$ $u_k(x)= 1+5cotx$ . How much work would it take to move the object over $x \in [\frac{π}{12}, \frac{3 π}{8}]$ $x in [(pi)/12, (3pi)/8]$ , where $x$ $x$ is in meters?

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An object with a mass of $6 k g$ $6 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 1 + 5 cot x$ $u_k(x)= 1+5cotx$ . How much work would it take to move the object over $x \in [\frac{π}{12}, \frac{3 π}{8}]$ $x in [(pi)/12, (3pi)/8]$ , where $x$ $x$ is in meters?

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Answer 1 · 2018-04-03T07:14:41

The work is $= 428.1 J$ $=428.1J$

Explanation:

$Reminder :$ $"Reminder : "$

$\int cot x d x = ln (| sin x |) + C$ $intcotxdx=ln(|sinx|)+C$

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 coefficient of kinetic friction is $μ_{k} = (1 + 5 cot x)$ $mu_k=(1+5cotx)$

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

The mass of the object is $m = 6 k g$ $m=6kg$

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

$= 6 \cdot (1 + 5 cot x) g$ $=6*(1+5cotx)g$

The work done is

$W = 6 g \int_{\frac{1}{12} π}^{\frac{3}{8} π} (1 + 5 cot) d x$ $W=6gint_(1/12pi)^(3/8pi)(1+5cot)dx$

$= 6 g \cdot [x + 5 {ln (sin (x)]}_{\frac{1}{12} π}^{\frac{3}{8} π}$ $=6g*[x+5ln(sin(x) ] _(1/12pi)^(3/8pi)$

$= 6 g ((\frac{3}{8} π + 5 ln (sin (\frac{3}{8} π)) - (\frac{1}{12} π + 5 ln (sin (\frac{1}{12} π))$ $=6g((3/8pi+5ln(sin(3/8pi))-(1/12pi+5ln(sin(1/12pi))$

$= 6 \times 9.8 \times 7.28$ $=6xx9.8xx7.28$

$= 428.1 J$ $=428.1J$

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An object with a mass of 6 kg6kg6 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 1+5cotx uk(x)=1+5cotxu_k(x)= 1+5cotx . How much work would it take to move the object over x in [(pi)/12, (3pi)/8]x∈[π12,3π8]x in [(pi)/12, (3pi)/8], where xxx is in meters?

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