Question

An object with a mass of `8 kg` is pushed along a linear path with a kinetic friction coefficient of `u_k(x)= 2x^2-3x`. How much work would it take to move the object over #x in [2, 3], where x is in meters?

Answer 1

`797.1J` rounded to one place of decimal

Explanation:

Force of kinetic friction which needs to be overcome to move the object

`F_k=`Coefficient of kinetic friction `mu_ktimes`normal force `eta`
where `eta=mg`
In the problem `mu_k` is stated to be `u_k(x)`
Inserting given quantities and taking the value of `g=9.8 m//s^2`
`F_k=(2x^2-3x)times 8times 9.8 N`
`F_k=78.4(2x^2-3x) N`

When this force moves through a small distance `dx`, the work done is given as
`F_k cdotdx=[78.4(2x^2-3x)]cdot dx`
When the force moves through a distance from `x in [2, 3]`, total work done is integral of RHS over the given interval.

Total work done`=int_2^3 78.4(2x^2-3x)cdotdx`
`implies`Total work done`=78.4 int_2^3 (2x^2-3x)cdotdx`

`=78.4 (2x^3 /3-3x^2/2+C)|_2^3`, where C is constant of integration.
`=78.4 [ (2 3^3 /3-3 3^2/2+cancel C)-(2 2^3 /3-3 2^2/2+cancel C)]`
`=78.4 [ 18-27/2-16/3+6]`
`=78.4 [ (108-51-32+36)/6]`
`=78.4 [ 61/6]`
`797.1J` rounded to one place of decimal