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

`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)}`.)

Step-by-step 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}`.