# An object with a mass of 4 kg is acted on by two forces. The first is F_1= < -2 N , -1 N> and the second is F_2 = < 5 N, 2 N>. What is the object's rate and direction of acceleration?

Aug 8, 2017

$\vec{a} = \left(0.79 {\text{m"//"s}}^{2} , {53}^{o}\right)$ or $< \frac{3}{4} , \frac{1}{4} > {\text{m"//"s}}^{2}$

#### Explanation:

$\textcolor{b l u e}{{\vec{F}}_{n e t} = m \vec{a}}$

where $m$ is the mass of the object and $\vec{a}$ is the object's acceleration, caused by net force ${\vec{F}}_{n e t}$

The acceleration is therefore given by:

$\textcolor{p u r p \le}{a = \frac{{F}_{n e t}}{m}}$

So, in order to calculate the acceleration, we'll have to find the net force first. This means finding the resultant force of the two forces provided.

We can express the acceleration as either a vector or by its magnitude.

Since both of the given forces are in vector notation, we can find ${F}_{n e t}$ as a vector using simple vector addition. We have:

• $\to {\vec{F}}_{1} = < - 2 \text{N",-1"N} >$

• $\to {\vec{F}}_{2} = < 5 \text{N",2"N} >$

Therefore:

${\vec{F}}_{n e t} = {\vec{F}}_{1} + {\vec{F}}_{2} = < \left(- 2 + 5\right) \text{N",(-1+2)"N} >$

$= < 3 \text{N",1"N} >$

And so we can calculate the acceleration as:

$\vec{a} = \left(< 3 \text{N",1"N">)/(4"kg}\right)$

$= < \frac{3}{4} , \frac{1}{4} > {\text{m"//"s}}^{2}$

The magnitude of the vector is then given by:

$\left\mid \vec{a} \right\mid = a = \sqrt{{a}_{x}^{2} + {a}_{y}^{2}}$

$= \sqrt{{\left(\frac{3}{4}\right)}^{2} + {\left(\frac{1}{4}\right)}^{2}}$

$\approx \textcolor{g r e e n}{0.79 {\text{m"//"s}}^{2}}$

The direction of the acceleration will occur in the direction of the net force.

We found the net force above to be ${\vec{F}}_{n e t} = < 3 , 4 >$. We can use the inverse tangent (arctangent) function to calculate the angle of the net force with the horizontal and consequently the direction of acceleration.

$\theta = \arctan \left(\frac{{F}_{y}}{{F}_{x}}\right)$

$= \arctan \left(\frac{4}{3}\right)$

$\approx \textcolor{red}{{53}^{o}}$