Question

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?

Answer 1

283.53J

Explanation:

First we will write the information provided
`mu_k=e^x+x`
M=`5 Kg`

we know that the work done by us to the 5 kg mass is
`W=F_(applied)*ds`

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

`W=(F_(applied)-f_(k))*dS`

here we need to move the block without acceleration so the applied force must be equal to the kinetic friction .
`F_(applied)=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*dx=mu_k*N*dx=int_1^2(e^x+x)*m*g*dx=283.53J`

Answer 2

The work is `=302.3J`

Explanation:

The work done is

`W=F*d`

The frictional force is

`F_r=mu_k*N`

`N=mg`

`F_r=mu_k*mg`

`=5(e^x+x)g`

The work done is

`W=5gint_(1)^(2)(e^x+x)dx`

`=5g*[e^x+x^2/2]_(1)^(2)`

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

`=5g(9.39-3.22)`

`=5g*6.17`

`=302.3J`