An object with a mass of 2 kg is acted on by two forces. The first is F_1= <-3 N , 1 N> and the second is F_2 = < 6 N, -2 N>. What is the object's rate and direction of acceleration?

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

Explanation:

HINT: The acceleration $a$ of a rigid body of mass $m$ under the application of force $F$ is given as follows

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

Jul 4, 2018

The rate of acceleration is $= 1.58 m {s}^{-} 2$ in the direction ${18.43}^{\circ}$ clockwise from the x-axis

Explanation:

The resultant of the $2$ forces is

$\vec{F} = {\vec{F}}_{1} + {\vec{F}}_{2}$

$= < - 3 , 1 > + < 6 , - 2 >$

$= < 3 , - 1 >$

According to Newton's Second Law

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

The mass is $m = 2 k g$

Therefore,

The acceleration is

$\vec{a} = \frac{\vec{F}}{m} = \frac{1}{2} \cdot < 3 , - 1 > = < \frac{3}{2} , - \frac{1}{2} >$

The rate of acceleration is

$| | \vec{a} | | = | | < \frac{3}{2} , - \frac{1}{2} > | | = \sqrt{{\left(\frac{3}{2}\right)}^{2} + {\left(- \frac{1}{2}\right)}^{2}}$

$= \sqrt{\frac{9}{4} + \frac{1}{4}}$

$= \frac{\sqrt{10}}{2}$

$= 1.58 m {s}^{-} 2$

The direction of acceleration is

$\theta = \arctan \left(- \frac{1}{3}\right) = {18.43}^{\circ}$ clockwise from the x-axis