# 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?

Apr 17, 2017

The work done is $= 65.7 J$

#### Explanation:

We need

$\int {e}^{\times} \mathrm{dx} = {e}^{x}$

The frictional force is

$F = {F}_{r} = {\mu}_{k} \cdot N$

The normal force is $N = m g$

So,

${F}_{r} = {\mu}_{k} \cdot m g$

$m = 1 k g$

But,

${\mu}_{k} \left(x\right) = {e}^{x} - x + 3$

The work done is

$W = {F}_{r} \cdot d = {\int}_{0}^{\frac{\pi}{12}} {\mu}_{k} m g \mathrm{dx}$

$= g {\int}_{1}^{2} \left({e}^{x} - x + 3\right) \mathrm{dx}$

$= g {\left[{e}^{x} - {x}^{2} / 2 + 3 x\right]}_{1}^{2}$

$= g \left(\left({e}^{2} - 2 + 6\right) - \left(e - \frac{1}{2} + 3\right)\right)$

$= g \cdot \left({e}^{2} - e + \frac{3}{2}\right)$

$= 6.17 g$

$= 65.7 J$