# 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 ]?

Mar 11, 2016

$\Delta K = \frac{65}{6}$

#### Explanation:

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

$m \cdot a = F$

$m \ddot{x} = 4 {x}^{2} + x$

Multiply both sides by $\dot{x}$

$m \ddot{x} \dot{x} = 4 {x}^{2} \dot{x} + x \dot{x}$

then integrate both sides

$\frac{1}{2} m {\dot{x}}^{2} = \frac{4}{3} {x}^{3} + \frac{1}{2} {x}^{2} + {v}_{1}$

We know that the kinetic energy is given by

$K = \frac{1}{2} m {\dot{x}}^{2} = \frac{4}{3} {x}^{3} + \frac{1}{2} {x}^{2} + {v}_{1}$

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

$\Delta K = \left(\frac{4}{3} {2}^{3} + \frac{1}{2} {2}^{2} + {v}_{1}\right) - \left(\frac{4}{3} {1}^{3} + \frac{1}{2} {1}^{2} + {v}_{1}\right) = \frac{65}{6}$

Mar 11, 2016

$\Delta K = \frac{65}{6}$

#### 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.

$W = {\int}_{1}^{2} F \mathrm{dx} = {\int}_{1}^{2} \left(4 {x}^{2} + x\right) \mathrm{dx} = {\left[\frac{4 {x}^{3}}{3} + {x}^{2} / 2\right]}_{1}^{2}$
$\text{ } = \frac{4 \cdot {2}^{3}}{3} + {2}^{2} / 2 - \frac{4 \cdot {1}^{3}}{3} - {1}^{2} / 2 = \frac{65}{6}$

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

$\Delta K = \frac{65}{6}$