Question

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

Answer 1

Work `=269.5J`

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`
Inserting given quantities and taking the value of `g=9.8 m//s^2`
`F_k=(x+3)times 5times 9.8 N`
`F_k=49(x+3) N`

When this force moves through a small distance `dx`, the work done is given as
`F_k dx=49(x+3) 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 49(x+3)dx`
`implies`Total work done`=49 int_2^3 (x+3)dx`

`=49 (x^2 /2+3x+C)|_2^3`, where C is constant of integration.
`=49 [ (3^2 /2+3times3+C)-(2^2 /2+3times2+C)]`
`=49 [ 9 /2+9-2-6]`
`=49 [ 4.5+1]`