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

May 7, 2017

283.53J

#### Explanation:

First we will write the information provided
${\mu}_{k} = {e}^{x} + x$
M=$5 K g$

we know that the work done by us to the 5 kg mass is
$W = {F}_{a p p l i e d} \cdot \mathrm{ds}$

but in this case there is friction is involved which can be included in the equation by ,

$W = \left({F}_{a p p l i e d} - {f}_{k}\right) \cdot \mathrm{dS}$

here we need to move the block without acceleration so the applied force must be equal to the kinetic friction .
${F}_{a p p l i e d} = {f}_{k}$

similarly the work done will also be the same in both the cases so the work done by the frictional force is given by .

$W = {f}_{k} \cdot \mathrm{dx} = {\mu}_{k} \cdot N \cdot \mathrm{dx} = {\int}_{1}^{2} \left({e}^{x} + x\right) \cdot m \cdot g \cdot \mathrm{dx} = 283.53 J$

May 7, 2017

The work is $= 302.3 J$

#### Explanation:

The work done is

$W = F \cdot d$

The frictional force is

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

$N = m g$

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

$= 5 \left({e}^{x} + x\right) g$

The work done is

$W = 5 g {\int}_{1}^{2} \left({e}^{x} + x\right) \mathrm{dx}$

$= 5 g \cdot {\left[{e}^{x} + {x}^{2} / 2\right]}_{1}^{2}$

=5g((e^2+2)-(e+1/2)))#

$= 5 g \left(9.39 - 3.22\right)$

$= 5 g \cdot 6.17$

$= 302.3 J$