# An object with a mass of 4 kg is acted on by two forces. The first is F_1= < -2 N , 4 N> and the second is F_2 = < 3 N, 5 N>. What is the object's rate and direction of acceleration?

Jun 24, 2017

$a = 2.26$ ${\text{m/s}}^{2}$

$\theta = {83.7}^{\text{o}}$

#### Explanation:

To solve this problem, we can use the equation

$\vec{a} = \frac{\Sigma \vec{F}}{m}$

or in the component equations

${a}_{x} = \frac{\Sigma {F}_{x}}{m}$

${a}_{y} = \frac{\Sigma {F}_{y}}{m}$

Our components for the forces are

${F}_{1 x} = - 2$ $\text{N}$

${F}_{2 x} = 3$ $\text{N}$

• $\Sigma {F}_{x} = 1$ $\text{N}$

${F}_{1 y} = 4$ $\text{N}$

${F}_{2 y} = 5$ $\text{N}$

• $\Sigma {F}_{y} = 9$ $\text{N}$

Using these two net force components, we can calculate the acceleration components using the equations above and it's mass ($4$ $\text{kg}$):

${a}_{x} = \left(1 \textcolor{w h i t e}{l} \text{N")/(4color(white)(l)"kg}\right) = 0.25$ ${\text{m/s}}^{2}$

${a}_{y} = \left(9 \textcolor{w h i t e}{l} \text{N")/(4color(white)(l)"kg}\right) = 2.25$ ${\text{m/s}}^{2}$

The magnitude of the acceleration is thus

a = sqrt((0.25"m"/("s"^2))^2 + (2.25"m"/("s"^2))^2) = color(red)(2.26 $\textcolor{red}{{\text{m/s}}^{2}}$

And it's direction is

theta = arctan((2.25"m"/("s"^2))/(0.25"m"/("s"^2))) = color(blue)(83.7^"o"