An object with a mass of $5 k g$ $5 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 1 - cos (\frac{x}{2})$ $u_k(x)= 1-cos(x/2)$ . How much work would it take to move the object over #x in [0, 8pi], where x is in meters?

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Answer 1 · 2018-03-09T07:11:10

The work is $= 1231.4 J$ $=1231.4J$

$Reminder :$ $"Reminder : "$

$\int cos x d x = sin x + C$ $intcosxdx=sinx+C$

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 $μ_{k} = (1 - cos (\frac{x}{2}))$ $mu_k=(1-cos(x/2))$

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

The mass of the object is $m = 5 k g$ $m=5kg$

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

$= 5 \cdot (1 - cos (\frac{x}{2})) g$ $=5*(1-cos(x/2))g$

The work done is

$W = 5 g \int_{0}^{8 π} (1 - cos (\frac{x}{2})) d x$ $W=5gint_(0)^(8pi)(1-cos(x/2))dx$

$= 5 g \cdot {[x - 2 sin (\frac{x}{2})]}_{0}^{8 π}$ $=5g*[x-2sin(x/2)]_(0)^(8pi)$

$= 5 g ((8 π - 2 \cdot 0) - (0 - 2 \cdot 0))$ $=5g((8pi-2*0)-(0-2*0))$

$= 5 g (8 π)$ $=5g(8pi)$

$= 1231.4 J$ $=1231.4J$

An object with a mass of 5 kg5kg5 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 1-cos(x/2) uk(x)=1−cos(x2)u_k(x)= 1-cos(x/2) . How much work would it take to move the object over #x in [0, 8pi], where x is in meters?