An object with a mass of $7 k g$ $7 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 1 - cos (\frac{x}{6})$ $u_k(x)= 1-cos(x/6)$ . 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 · 2018-01-04T06:53:55

The work is $= 2078.6 J$ $=2078.6J$

$Reminder :$ $"Reminder : "$

$\int cos a x d x = \frac{1}{a} sin a x + C$ $intcosaxdx=1/asinax+C$

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 $μ_{k} = (1 - cos (\frac{x}{6}))$ $mu_k=(1-cos(x/6))$

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

The mass of the object is $m = 7 k g$ $m=7kg$

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

$= 7 \cdot (1 - cos (\frac{x}{6})) g$ $=7*(1-cos(x/6))g$

The work done is

$W = 7 g \int_{0}^{8 π} (1 - cos (\frac{x}{6})) d x$ $W=7gint_(0)^(8pi)(1-cos(x/6))dx$

$= 7 g \cdot {[x - 6 sin (\frac{1}{6} x)]}_{0}^{8 π}$ $=7g*[x-6sin(1/6x)]_(0)^(8pi)$

$= 7 g (8 π - 6 sin (\frac{4}{3} π) - (0))$ $=7g(8pi-6sin(4/3pi)-(0))$

$= 7 g (30.3)$ $=7g(30.3)$

$= 2078.6 J$ $=2078.6J$

An object with a mass of 7kg7 kg is pushed along a linear path with a kinetic friction coefficient of uk(x)=1−cos(x6)u_k(x)= 1-cos(x/6) . How much work would it take to move the object over #x in [0, 8pi], where x is in meters?