Question

13. An object of mass $30 \mathrm{~kg}$ is falling in air and experiences a force due to air resistance of 50 newtons.
a. Determine the net force acting on the object and
b. calculate the accelerâtion of the object.

Answer 1

Answer:

$250\; {\rm N}$ (downwards.)

Approximately $8.3\; {\rm m\cdot s^{-2}}$

(Assuming that the gravitational field strength is $g \approx 10\; {\rm m\cdot s^{-2}}$.)

Explanation:

Note that $1\; {\rm kg \cdot m\cdot s^{-2}} = 1\; {\rm N}$.

There are two forces on this object:

• weight (downward), and
• air resistance (upwards.)

Let $g$ denote the gravitational field strength. The weight of an object of mass $m$ will be $m\, g$.

In this example, since $m = 30\; {\rm kg}$ and $g \approx 10\; {\rm m\cdot s^{-2}}$ around the surface of the earth. The weight of this object will be:

$\begin{aligned}m\, g &\approx (30\; {\rm kg})\, (10\; {\rm m\cdot s^{-2}}) \\ &= 300\; {\rm kg \cdot m\cdot s^{-2}} \\ &= 300\; {\rm N}\end{aligned}$.

The air resistance on this object is given to be $50\; {\rm N}$ (upwards.) Since the two forces are in opposite directions, the magnitude of the resultant force on the object will be the difference between their magnitudes:

$\begin{aligned} &(\text{resultant force}) \\ &= (\text{weight}) - (\text{resistance}) \\ & \approx (300\; {\rm N}) - (50\; {\rm N}) \\ &= 250\; {\rm N}\end{aligned}$.

(Downwards.)

Divide the resultant force on the object by the mass $m$ of the object to find the acceleration of the object:

$\begin{aligned}& (\text{acceleration}) \\ &= \frac{(\text{resultant force})}{(\text{mass})} \\ &\approx \frac{250\; {\rm N}}{30\; {\rm kg}} \\ &= 8.3\; {\rm m \cdot s^{-2}}\end{aligned}$.