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

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Answer 1 · 2017-05-01T08:24:28

The work is $= 146.7 J$ $=146.7J$

The work done is

$W = F \cdot d$ $W=F*d$

The frictional force is

$F_{r} = μ_{k} \cdot N$ $F_r=mu_k*N$

$N = m g$ $N=mg$

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

$= 6 (1 + 3 cot x) g$ $=6(1+3cotx)g$

The work done is

$W = \int_{\frac{π}{6}}^{\frac{3}{8} π} 6 (1 + 3 cot x) g d x$ $W=int_(pi/6)^(3/8pi)6(1+3cotx)gdx$

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

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

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

$= 6 g (\frac{5}{24} π + 1.84)$ $=6g(5/24pi+1.84)$

$= 6 g \cdot 2.49$ $=6g*2.49$

$= 146.7 J$ $=146.7J$

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