Question

The force applied against an object moving horizontally on a linear path is described by `F(x)= 4x^2+x`. By how much does the object's kinetic energy change as the object moves from `x in [ 1, 2 ]`?

Answer 1

`Delta K= 65/6`

Explanation:

We begin by writing the acceleration in terms of the force,

`m*a=F`

`m ddot x=4x^2+x`

Multiply both sides by `dot x`

`m ddot x dot x = 4x^2 dot x + x dot x`

then integrate both sides

`1/2 m dot x ^2=4/3 x^3 + 1/2 x^2+v_1`

We know that the kinetic energy is given by

`K=1/2 m dot x^2 = 4/3 x^3 + 1/2 x^2+v_1`

The change in kinetic energy will be the difference at the two point in `x`

`Delta K=(4/3 2^3 + 1/2 2^2+v_1)-(4/3 1^3 + 1/2 1^2+v_1) = 65/6`

Answer 2

`Delta K = 65/6`

Explanation:

I've decided that this is actually much easier than the previous answer. Let's use the concept of work which is the force integrated over the distance that is applied.

`W = int_1^2 F dx = int_1^2 (4x^2+x) dx = [(4x^3)/3+x^2/2]_1^2`
`" " = (4*2^3)/3+2^2/2-(4*1^3)/3-1^2/2 = 65/6`

The work that we do ends up increasing the kinetic energy of the object, so

`Delta K = 65/6`