Chapter SH, Problem 3.6APP

To determine

To Calculate: The force of air resistance on the object when it reaches terminal velocity.

Answer to Problem 3.6APP

  $24.5\text{N}$ (upwards)

Explanation of Solution

Given:

Mass of the object, $m=2.5\text{kg}$

Height from which the object drops, $h=1000\text{m}$

Formula Used:

From Newton’s second law of motion, the net force would be:

  ${\displaystyle \sum F=ma}$

Here, m is the mass and a is the acceleration.

Weight of an object can be obtained by:

  $w=mg$

Here, g is the acceleration due to gravity.

Calculations:

At terminal velocity, the acceleration of the object is zero. This means that the net force on the object is zero.

The forces acting on the object when it is in air are its weight acting in downward direction and the force of air resistance acting in opposite direction.

  $\begin{array}{l}{F}_{air}+w=0\\ F_{air}=-w\\ F_{air}=-mg\\ F_{air}=\text{-}\left(\text{2}\text{.5kg}\right)\left(\text{9}{\text{.8m/s}}^{\text{2}}\right)\\ F_{air}=-24.5\text{N}\end{array}$

Conclusion:

Thus, the force of air resistance is $24.5\text{N}$ (upwards).