An object with a mass of $7 k g$ $7 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_{k} (x) = 1 + x^{2} - x cos (x)$ $u_k(x)= 1+x^2-xcos(x)$ . How much work would it take to move the object over #x in [pi, 4pi], where x is in meters?

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Answer 1 · 2018-02-22T13:02:30

The work done is $= 45.18 k J$ $=45.18kJ$

$Reminder :$ $"Reminder : "$

Integraton by parts

$intuv'=uv-intu'v$

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

$v'=cosx$ , $=>$ , $v=sinx$

$intxcosxdx=xsinx+cosx+C$

The work done is

$W=F*d$

The frictional force is

$F_r=mu_k*N$

The coefficient of kinetic friction is $mu_k=(1+x^2-xcos(x))$

The normal force is $N=mg$

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

$F_r=mu_k*mg$

$=7*(1+x^2-xcos(x))g$

The work done is

$W=7gint_(pi)^(4pi)(1+x^2-xcos(x))dx$

$=7g*[x+1/3x^3-xsin(x)-cosx]_(pi)^(4pi)$

$=7g(4pi+64/3pi^3-4pisin(4pi)-cos(4pi))-(pi+1/3pi^3+1))$

$=7g(-2+3pi+21pi^3)$

$=45177.2J$

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+x^2-xcos(x) uk(x)=1+x2−xcos(x)u_k(x)= 1+x^2-xcos(x) . How much work would it take to move the object over #x in [pi, 4pi], where x is in meters?