# A 3-kg rock is thrown upward with a force 2

A 3-kg rock is thrown upward with a force of 200 N at a location where the local gravitational acceleration is 9.79 $m/s^2$. Determine the acceleration of the rock, in $m/s^2$.

#### Solution:

Let us first calculate the weight of the rock using the following equation:

$W=mg$

(Where $W$ is weight, $m$ is mass and $g$ is acceleration due to gravity)

Substitute the values we know:

$W=mg$

$W=(3)(9.79)$

$W=29.37$ N

We will now calculate all the forces that affect the rock (net force):

$\uparrow^+ F_{net}=200-29.37$

$F_{net}=170.63$ N

Let us now calculate the acceleration using Newtons second law.

$F=ma$

(Where $F$ is force, $m$ is mass and $a$ is acceleration)

Substituting the values we know gives us:

$F=ma$

(isolate for $a$)

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

$a=\dfrac{170.63}{3}$

$a=56.88\,\text{m}/\text{s}^2$

$a=56.88\,\text{m}/\text{s}^2$