# An object with a mass of 4 kg is traveling at 1 m/s. If the object is accelerated by a force of f(x) = x^2 -x +1  over x in [1, 9], where x is in meters, what is the impulse at x = 2?

Feb 8, 2016

Impulse $J = \setminus \Delta p = 1.5377 N . s$

#### Explanation:

Impulse-Momentum Theorem: $\setminus \vec{{J}_{}} = \setminus \Delta \setminus \vec{{p}_{}}$,

To calculate the Impulse ( $\setminus \vec{J}$) just calculate the change in momentum ( $\setminus \Delta \setminus \vec{{p}_{}}$ ). To calculate the change in momentum, remember the relation between momentum ($p$) and the kinetic energy ($K$) :
$K = {p}^{2} / \left(2 m\right) \setminus \rightarrow p = \setminus \sqrt{2 m K}$
Therefore $\setminus \Delta p = \setminus \sqrt{2 m} \left(\setminus \sqrt{{K}_{f}} - \setminus \sqrt{{K}_{i}}\right)$

Work-Energy Theorem: The total work done by all the forces acting on the object must be equal to the change in its kinetic energy $W = \setminus \Delta K = {K}_{f} - {K}_{i}$, where ${K}_{i}$ and ${K}_{f}$ are the initial and final kinetic energies.
${K}_{f} = {K}_{i} + W$

${K}_{i} = \frac{1}{2} m {v}_{i}^{2} = \frac{1}{2} \left(4 k g\right) {\left(1 \frac{m}{s}\right)}^{2} = 2$ Joules,
$W = \setminus {\int}_{1}^{2} F \left(x\right) \mathrm{dx} = \setminus {\int}_{1}^{2} \left({x}^{2} - x + 1\right) \mathrm{dx} = {\left[{x}^{3} / 3 - {x}^{2} / 2 + x\right]}_{1}^{2}$
$\setminus q \quad \setminus \quad$$= \left({2}^{3} / 3 - {2}^{2} / 2 + 2\right) - \left(\frac{1}{3} - \frac{1}{2} + 1\right) = \frac{11}{6}$ Joules
${K}_{f} = {K}_{i} + W = 2 + \frac{11}{6} = \frac{23}{6}$ Joules

Now we are ready to evaluate the impulse:
$J = \setminus \Delta p = \setminus \sqrt{2 m} \left(\setminus \sqrt{{K}_{f}} - \setminus \sqrt{{K}_{i}}\right) = 1.5377 N . s$