# An object with a mass of 2 kg is acted on by two forces. The first is F_1= <4 N , 8 N> and the second is F_2 = < 1 N, 7 N>. What is the object's rate and direction of acceleration?

Jan 4, 2017

The objects rate of acceleration is $7.91 \setminus m {s}^{-} 2$ (2dp)
The direction of the acceleration is at an angle of 71.6º (1dp)

#### Explanation:

Let the resultant force acting on the object be $\vec{F}$, Then

$\vec{F} = {\vec{F}}_{1} + {\vec{F}}_{2}$
$\setminus \setminus \setminus \setminus = \left\langle4 , 8\right\rangle + \left\langle1 , 7\right\rangle$
$\setminus \setminus \setminus \setminus = \left\langle5 , 15\right\rangle$

The magnitude, $F$ of $\vec{F}$ is given by it's norm;

$F = | \vec{F} |$
$\setminus \setminus \setminus \setminus = | \left\langle5 , 15\right\rangle |$
$\setminus \setminus \setminus \setminus = \sqrt{{5}^{2} + {15}^{2}}$
$\setminus \setminus \setminus \setminus = \sqrt{25 + 225}$
$\setminus \setminus \setminus \setminus = \sqrt{250}$
$\setminus \setminus \setminus \setminus = 15.81 \setminus N$ (2dp)

Applying Newton's 2nd Law of Motion, $F = m a$, we have:

$\therefore 15.81 = 2 a$
$\therefore a = 7.91 \setminus m {s}^{-} 2$ (2dp)

The direction of the acceleration is the same as the direction of the resultant force $\vec{F}$. If this is at an angle $\theta$ then

$\tan \theta = \frac{15}{5}$
$\therefore \setminus \theta = \arctan 3$
 :. \ theta = 71.6º (1dp)

Hence,

The objects rate of acceleration is $7.91 \setminus m {s}^{-} 2$ (2dp)
The direction of the acceleration is at an angle of 71.6º