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

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Answer 1 · 2018-07-24T09:20:51

The work is $= 174.64 J$ $=174.64J$

$Reminder :$ $"Reminder : "$

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

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

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

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

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

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

The work done is

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

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

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

$= 6 g (2.97)$ $=6g(2.97)$

$= 174.64 J$ $=174.64J$

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+cotx uk(x)=1+cotxu_k(x)= 1+cotx . How much work would it take to move the object over #x in [(pi)/12, (5pi)/8], where x is in meters?