An object with a mass of $7 k g$ $7 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 1 + x cos (\frac{π}{4} - \frac{x}{6})$ $u_k(x)= 1+xcos(pi/4-x/6)$ . How much work would it take to move the object over #x in [0, 4pi], where x is in meters?

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Answer 1 · 2017-07-13T14:44:17

The work is $= 494.6 J$ $=494.6J$

The integration by parts is

$intuv'dx=uv-intu'vdx$

$F_r=mu_k*N$

$dW=F_r*dx$

We start by calculating the integral of

$intxcos(pi/4-x/6)$

We perform this by integration by parts

$u=x$ , $=>$ , $u'=1$

$v'=cos(pi/4-x/6)$ , $=>$ , $v=-6sin(pi/4-x/6)$

Therefore,

$intxcos(pi/4-x/6)=-6xsin(pi/4-x/6)+int1*6sin(pi/4-x/6)$

$=-6xsin(pi/4-x/6)+36cos(pi/4-x/6)$

The work is

$W=7gint_0^(4pi)(1+xsin(pi/4-x/6))$

$=7g[x-6xsin(pi/4-x/6)+36cos(pi/4-x/6)]_0^(4pi)$

$=7g((4pi-24pisin(-5/12pi)+36cos(-5/12pi))-(0-0+36cos(pi/4)))$

$=7g(69.259)$

$=494.6J$

An object with a mass of 7 kg7kg7 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 1+xcos(pi/4-x/6) uk(x)=1+xcos(π4−x6)u_k(x)= 1+xcos(pi/4-x/6) . How much work would it take to move the object over #x in [0, 4pi], where x is in meters?