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 $x \in [\frac{5 π}{12}, \frac{5 π}{8}]$ $x in [(5pi)/12, (5pi)/8]$ , 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 $x \in [\frac{5 π}{12}, \frac{5 π}{8}]$ $x in [(5pi)/12, (5pi)/8]$ , where $x$ $x$ is in meters?

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Answer 1 · 2018-02-25T12:41:21

The work is $= 25.28 J$ $=25.28J$

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 $μ_{k} = (1 + 5 cot (x))$ $mu_k=(1+5cot(x))$

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

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

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

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

The work done is

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

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

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

$= 6 g (\cdot 0.43)$ $=6g(*0.43)$

$= 25.28 J$ $=25.28J$

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An object with a mass of 6kg6 kg is pushed along a linear path with a kinetic friction coefficient of uk(x)=1+5cotxu_k(x)= 1+5cotx. How much work would it take to move the object over x∈[5π12,5π8]x in [(5pi)/12, (5pi)/8], where xx is in meters?

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