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+xsinx`. How much work would it take to move the object over #x in [0, 5pi], where x is in meters?

Answer 1

`7 "kJ"`

Explanation:

The weight of the object is given by `W = m*g`. Since the object is not accelerating in the vertical direction, the normal force, `N`, is exactly equal to the weight.

The friction is given by `F = u_"k" * N`.

Consider the object being dragged/pushed along a small distance `"d"x`. The amount of work needed is `F * "d"x = N u_k(x) "d"x`.

The total work done to oppose the friction is given by

`int_0^{5pi} F dx = N int_0^{5pi} u_k(x) "d"x`

`= mg int_0^{5pi} (x+xsin(x)) "d"x qquad` (integrate by parts)

`= mg [x^2/2 - xcos(x) + sin(x)]_0^{5pi}`

`= mg (frac{25pi^2}{2} + 5pi)`

`= (5 "kg") (9.81 "m/s"^2) (frac{25pi^2}{2} + 5pi)`

`= 6821 "J"`