# The force applied against an object moving horizontally on a linear path is described by F(x)=x^2-3x + 3 . By how much does the object's kinetic energy change as the object moves from  x in [ 0 , 1 ]?

Apr 13, 2016

Newton's second law of motion:

$F = m \cdot a$

Definitions of acceleration and velocity:

$a = \frac{\mathrm{du}}{\mathrm{dt}}$

$u = \frac{\mathrm{dx}}{\mathrm{dt}}$

Kinetic energy:

$K = m \cdot {u}^{2} / 2$

ΔK=11/6 $k g \cdot {m}^{2} / {s}^{2}$

#### Explanation:

Newton's second law of motion:

$F = m \cdot a$

${x}^{2} - 3 x + 3 = m \cdot a$

Substituting $a = \frac{\mathrm{du}}{\mathrm{dt}}$ does not help with the equation, since $F$ isn't given as a function of $t$ but as a function of $x$ However:

$a = \frac{\mathrm{du}}{\mathrm{dt}} = \frac{\mathrm{du}}{\mathrm{dt}} \cdot \frac{\mathrm{dx}}{\mathrm{dx}} = \frac{\mathrm{dx}}{\mathrm{dt}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

But $\frac{\mathrm{dx}}{\mathrm{dt}} = u$ so:

$a = \frac{\mathrm{dx}}{\mathrm{dt}} \cdot \frac{\mathrm{du}}{\mathrm{dx}} = u \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Substituting into the equation we have, we have a differential equation:

${x}^{2} - 3 x + 3 = m \cdot u \frac{\mathrm{du}}{\mathrm{dx}}$

$\left({x}^{2} - 3 x + 3\right) \mathrm{dx} = m \cdot u \mathrm{du}$

${\int}_{{x}_{1}}^{{x}_{2}} \left({x}^{2} - 3 x + 3\right) \mathrm{dx} = {\int}_{{u}_{1}}^{{u}_{2}} m \cdot u \mathrm{du}$

The two speeds are unknown but the positions $x$ are known. Also, mass is constant:

${\int}_{0}^{1} \left({x}^{2} - 3 x + 3\right) \mathrm{dx} = m \cdot {\int}_{{u}_{1}}^{{u}_{2}} u \mathrm{du}$

${\left[{x}^{3} / 3 - 3 {x}^{2} / 2 + 3 x\right]}_{0}^{1} = m \cdot {\left[{u}^{2} / 2\right]}_{{u}_{1}}^{{u}_{2}}$

$\left({1}^{3} / 3 - 3 \cdot {1}^{2} / 2 + 3 \cdot 1\right) - \left({0}^{3} / 3 - 3 \cdot {0}^{2} / 2 + 3 \cdot 0\right) = m \cdot \left({u}_{2}^{2} / 2 - {u}_{1}^{2} / 2\right)$

$\frac{11}{6} = m \cdot {u}_{2}^{2} / 2 - m \cdot {u}_{2}^{2} / 2$

But $K = m \cdot {u}^{2} / 2$

$\frac{11}{6} = {K}_{2} - {K}_{1}$

ΔK=11/6 $k g \cdot {m}^{2} / {s}^{2}$

Note : the units are $k g \cdot {m}^{2} / {s}^{2}$ only if the distances given $\left(x \in \left[0 , 1\right]\right)$ are in metres.