An object with a mass of $6 k g$ $6 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 1 + csc x$ $u_k(x)= 1+cscx$ . How much work would it take to move the object over #x in [(3pi)/4, (7pi)/8], where x is in meters?

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Answer 1 · 2018-05-01T18:04:20

The work is $= 66.21 J$ $=66.21J$

$Reminder :$ $"Reminder : "$

$\int csc x d x = ln ∣ ∣ (tan (\frac{x}{2})) ∣ ∣ + C$ $intcscxdx=ln|(tan(x/2))|+C$

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

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

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

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

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

The work done is

$W = 6 g \int_{\frac{3}{4} π}^{\frac{7}{8} π} (1 + csc (x)) d x$ $W=6gint_(3/4pi)^(7/8pi)(1+csc(x))dx$

$= 6 g \cdot {[x + ln ∣ ∣ (tan (\frac{x}{2})) ∣ ∣]}_{\frac{3}{4} π}^{\frac{7}{8} π}$ $=6g*[x+ln|(tan(x/2))|]_(3/4pi)^(7/8pi)$

$= 6 g (\frac{7}{8} π + ln (tan (\frac{7}{16} π))) - (\frac{3}{4} π + ln (tan (\frac{3}{8} π))$ $=6g(7/8pi+ln(tan(7/16pi)))-(3/4pi+ln(tan(3/8pi))$

$= 6 g (1.126)$ $=6g(1.126)$

$= 66.21 J$ $=66.21J$

An object with a mass of 6kg6 kg is pushed along a linear path with a kinetic friction coefficient of uk(x)=1+cscxu_k(x)= 1+cscx . How much work would it take to move the object over #x in [(3pi)/4, (7pi)/8], where x is in meters?