An object with a mass of $5 k g$ $5 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 1 + sin (\frac{x}{3})$ $u_k(x)= 1+sin(x/3)$ . 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 · 2017-06-14T05:25:36

The work is $= 1452 J$ $=1452J$

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 $m = 5 k g$ $m=5kg$

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

$= 5 (1 + sin (\frac{x}{3})) g$ $=5(1+sin(x/3))g$

The work done is

$W = 5 g \int_{0}^{8 π} (1 + sin (\frac{x}{3})) d x$ $W=5gint_(0)^(8pi)(1+sin(x/3))dx$

$= 5 g \cdot {[x - 3 cos (\frac{x}{3})]}_{0}^{8 π}$ $=5g*[x-3cos(x/3)]_(0)^(8pi)$

$= 5 g ((8 π + \frac{3}{2}) - (0 - 3))$ $=5g((8pi+3/2)-(0-3))$

$= 5 g (8 π + \frac{9}{2})$ $=5g(8pi+9/2)$

$= 5 g (29.63)$ $=5g(29.63)$

$= 1452 J$ $=1452J$

An object with a mass of 5kg5 kg is pushed along a linear path with a kinetic friction coefficient of uk(x)=1+sin(x3)u_k(x)= 1+sin(x/3) . How much work would it take to move the object over #x in [0, 8pi], where x is in meters?