An object with a mass of $6 k g$ $6 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 2 + csc x$ $u_k(x)= 2+cscx$ . How much work would it take to move the object over #x in [pi/8, (3pi)/8], where x is in meters?

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Answer 1 · 2018-01-21T06:55:26

The work is $= 163.5 J$ $=163.5J$

$Reminder :$ $"Reminder : "$

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

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 $μ_{k} = (2 + csc (x))$ $mu_k=(2+csc(x))$

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

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

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

$= 6 \cdot (2 + csc (x)) g$ $=6*(2+csc(x))g$

The work done is

$W = 6 g \int_{\frac{1}{8} π}^{\frac{3}{8} π} (2 + csc (x)) d x$ $W=6gint_(1/8pi)^(3/8pi)(2+csc(x))dx$

$= 6 g \cdot {[2 x + ln ∣ ∣ (tan (\frac{x}{2})) ∣ ∣]}_{\frac{1}{8} π}^{\frac{3}{8} π}$ $=6g*[2x+ln|(tan(x/2))|]_(1/8pi)^(3/8pi)$

$= 6 g (\frac{3}{4} π + ln (tan (\frac{3}{16} π))) - (\frac{1}{4} π + ln (tan (\frac{1}{16} π))$ $=6g(3/4pi+ln(tan(3/16pi)))-(1/4pi+ln(tan(1/16pi))$

$= 6 g (2.78)$ $=6g(2.78)$

$= 163.5 J$ $=163.5J$

An object with a mass of 6kg6 kg is pushed along a linear path with a kinetic friction coefficient of uk(x)=2+cscxu_k(x)= 2+cscx . How much work would it take to move the object over #x in [pi/8, (3pi)/8], where x is in meters?