An object with a mass of 5 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 6x-1 . How much work would it take to move the object over #x in [2, 3], where x is in meters?

Jan 19, 2016

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Explanation:

The question is similar to another question solved as bellow. The difference is in the value of coefficient of friction. Choose appropriate value and solve.
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Force of kinetic friction which needs to be overcome to move the object

${F}_{k} =$Coefficient of kinetic friction ${\mu}_{k} \times$normal force $\eta$
where $\eta = m g$
Inserting given quantities and taking the value of $g = 9.8 m / {s}^{2}$
${F}_{k} = \left(x + 3\right) \times 5 \times 9.8 N$
${F}_{k} = 49 \left(x + 3\right) N$

When this force moves through a small distance $\mathrm{dx}$, the work done is given as
${F}_{k} \mathrm{dx} = 49 \left(x + 3\right) \mathrm{dx}$
When the force moves through a distance from $x \in \left[2 , 3\right]$, total work done is integral of RHS over the given interval.

Total work done$= {\int}_{2}^{3} 49 \left(x + 3\right) \mathrm{dx}$
$\implies$Total work done$= 49 {\int}_{2}^{3} \left(x + 3\right) \mathrm{dx}$

$= 49 \left({x}^{2} / 2 + 3 x + C\right) {|}_{2}^{3}$, where C is constant of integration.
$= 49 \left[\left({3}^{2} / 2 + 3 \times 3 + C\right) - \left({2}^{2} / 2 + 3 \times 2 + C\right)\right]$
$= 49 \left[\frac{9}{2} + 9 - 2 - 6\right]$
$= 49 \left[4.5 + 1\right]$