An object with a mass of $1 kg$ is pushed along a linear path with a kinetic friction coefficient of $u_k(x)= e^x-x+3$ . How much work would it take to move the object over #x in [1, 2], where x is in meters?

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Answer 1 · 2017-04-17T12:02:49

The work done is $=65.7J$

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We need

$inte^xxdx=e^x$

The frictional force is

$F=F_r=mu_k*N$

The normal force is $N=mg$

So,

$F_r=mu_k*mg$

$m=1kg$

But,

$mu_k(x)=e^x-x+3$

The work done is

$W=F_r*d=int_0^(pi/12)mu_kmgdx$

$=gint_1^2(e^x-x+3)dx$

$=g[e^x-x^2/2+3x]_1^2$

$=g((e^2-2+6)-(e-1/2+3))$

$=g*(e^2-e+3/2)$

$=6.17g$

$=65.7J$

An object with a mass of 1 kg1 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= e^x-x+3 u_k(x)= e^x-x+3 . How much work would it take to move the object over #x in [1, 2], where x is in meters?