# What is the net force acting on a falling 1-kg ball if it encounters 2N of air resistance?

Oct 18, 2015

$\textcolor{g r e e n}{\text{Net Force"=-7.8"N}}$

#### Explanation:

Start by computing for the weight of the ball.

$\left[1\right] \textcolor{w h i t e}{X X} w = m g$

$\left[2\right] \textcolor{w h i t e}{X X} w = \left(1 {\text{kg")(9.8"m/s}}^{2}\right)$

$\left[3\right] \textcolor{w h i t e}{X X} w = 9.8 {\text{kg"*"m/s}}^{2}$

$\left[4\right] \textcolor{w h i t e}{X X} \textcolor{red}{w = 9.8 \text{N}}$

Now let's create a diagram representing the forces acting on the ball:

$\textcolor{w h i t e}{X X} B A L L$
$\downarrow \textcolor{red}{9.8 \text{N}}$ $\uparrow \textcolor{b l u e}{2 \text{N}}$

Simply add the two forces together to get the net force. The weight $\textcolor{red}{9.8 \text{N}}$ is negative since it is directed downwards, while the air resistance $\textcolor{b l u e}{2 \text{N}}$ is positive since it is directed upwards.

"Net Force"=color(red)(-9.8"N")+color(blue)(2"N")

$\textcolor{g r e e n}{\text{Net Force"=-7.8"N}}$