An object with a mass of $7 k g$ $7 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 1 - cos (\frac{π}{4} - \frac{x}{6})$ $u_k(x)= 1-cos(pi/4-x/6)$ . 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 · 2017-11-26T07:08:15

The work is $= 1539.6 J$ $=1539.6J$

We need to calculate the integral $\int cos (\frac{1}{4} π - \frac{x}{6}) d x$ $intcos(1/4pi-x/6)dx$

Perform this integration by substitution

Let $u = \frac{1}{4} π - \frac{x}{6}$ $u=1/4pi-x/6$ , $\Rightarrow$ $=>$ , $d u = - \frac{1}{6} d x$ $du=-1/6dx$

$\int cos (\frac{1}{4} π - \frac{x}{6}) d x = - 6 \int cos u d u$ $intcos(1/4pi-x/6)dx=-6intcosudu$

$= - 6 sin u = - 6 sin (\frac{1}{4} π - \frac{x}{6})$ $=-6sinu=-6sin(1/4pi-x/6)$

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

The mass is $m = 7 k g$ $m=7kg$

The acceleration due to gravity is $g = 9.8 m s^{- 2}$ $g=9.8ms^-2$

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

$= 7 \cdot (1 - cos (\frac{π}{4} - \frac{x}{6})) g$ $=7*(1-cos(pi/4-x/6))g$

The work done is

$W = 7 g \int_{0}^{8 π} (1 - cos (\frac{π}{4} - \frac{x}{6})) d x$ $W=7gint_(0)^(8pi)(1-cos(pi/4-x/6))dx$

$= 7 g {[x + 6 sin (\frac{π}{4} - \frac{x}{6})]}_{0}^{8 π}$ $=7g[x+6sin(pi/4-x/6)]_0^(8pi)$

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

$= (7 \times g \times 22.44) J$ $=(7xxgxx22.44)J$

$= 1539.6 J$ $=1539.6J$

An object with a mass of 7kg7 kg is pushed along a linear path with a kinetic friction coefficient of uk(x)=1−cos(π4−x6)u_k(x)= 1-cos(pi/4-x/6) . How much work would it take to move the object over #x in [0, 8pi], where x is in meters?