Question

The forces in FIGURE EX$6.8$act on a $2.0kg$object. What are the values of $a_x$and $a_y$, the x- and y-components of the object’s acceleration?

Short answer

(a) Acceleration of the object in x-direction $a_x$= $-1m/s^2$.

(b) Acceleration of the object in y-direction $a_y$= $0m/s^2$.

Step-by-step solution

Acceleration : 

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

(a) Explanation : 

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,

$m{a}_X=4.0N-2.0N$

Substitute $2.0kg$for $m$to find $a_X$.

$$\begin{gathered} \left(2.0kg\right)a_X=2.0N-4.0N \\ {a}_X=\left(\frac{-2.0N}{2.0kg}\right)\left(\frac{1kg·m/s^2}{1N}\right) \\ =-1m/s^2 \end{gathered}$$

The x-component of acceleration acts in the negative x-direction, as shown by the negative sign.

Hence, acceleration of the object in x-direction =$-1m/s^2$.

(b) Explanation : 

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,

$m{a}_y=3.0N-3.0N$

Substitute $2.0kg$for $m$to find $a_y$.

$$\begin{gathered} \left(2.0kg\right)a_y=3.0N-3.0N \\ {a}_y=0m/s^2 \end{gathered}$$

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