The forces in FIGURE EX$6.9$act on a $2.0$kg object. What are the values of $a_X$and $a_y$, the x- and y-components of the object’s acceleration?

Acceleration is determined by the rate at which the velocity of an object changes over time.

Object's mass = $2.0kg$.

In the horizontal direction, Newton's second law applies,

$\sum _{}{F}_X=m{a}_X$

$m$= object's mass,

$a_X$= object's acceleration in x-direction,

In the x-direction, resolving the components of forces,

$m{a}_X=5.0N-1.0N-\left(3.0N\right)\left(\sin \theta \right)$

Substitute $2.0kg$for $m$and $20°$for $\theta$to find $a_X$.

$$\begin{gathered} \left(2.0kg\right)a_X=5.0N-1.0N-\left(3.0N\right)\left(\sin 20°\right) \\ \left(2.0kg\right)a_X=4.0N-\left(3.0N\right)\left(0.342\right) \\ =\left(\frac{2.98N}{2.0kg}\right)\left(\frac{1kg·m/s^2}{1N}\right) \\ \approx 1.5m/s^2 \end{gathered}$$

Hence, acceleration of object in x-direction is$1.5m/s^2$.

Object's mass = $2.0kg$.

In the vertical direction, Newton's second law applies,

$\sum _{}{F}_y=m{a}_y$

$m$= object's mass,

$a_y$= object's acceleration in y-direction,

In the y-direction, resolving the components of forces,

$m{a}_y=2.82N-(3.0N)\left(\cos \theta \right)$

Substitute $2.0kg$for $m$and $20°$for $\theta$to find $a_y$.

$$\begin{gathered} \left(2.0kg\right)a_y=2.82N-\left(3.0N\right)\left(\cos 20°\right) \\ =2.82N-\left(3.0N\right)\left(0.939\right) \\ =2.82N-2.82N \\ =0m/s^2 \end{gathered}$$

Hence, acceleration of the object in y-direction is$0m/s^2$.