Chapter 6, Problem 64A

A falling 50-kg parachutist experiences an upward acceleration of $6.2{\text{ m/s}}^{\text{2}}$ when she opens her parachute. Show that the drag force is 810 N when this occurs.

To determine

To show: The drag force is 810 N.

Explanation of Solution

Given:

Mass of parachutist, $m=50\text{kg}$

Upward acceleration, $a=6.2{\text{m/s}}^2$

Drag force, $F_{\text{drag}}=810\text{N}$

Formula Used:

From Newton’s second law,

Net force $F=ma$

Where, $m$ is mass of the object and $a$ is the acceleration.

Calculation:

The net force on the parachutist is,

  $\begin{array}{l}{F}_{\text{net}}=F_{\text{drag}}-mg\\ \Rightarrow F_{\text{drag}}=ma+mg\left(\text{since, }{F}_{\text{net}}=ma\right)\\ \Rightarrow F_{\text{drag}}=m\left(a+g\right)\end{array}$

Taking, $g=10{\text{m/s}}^2$ and substituting in above equation

  $\begin{array}{l}{F}_{\text{drag}}=50\left(6.2+10\right)\\ =50\times 16.2\\ =810\text{N}\end{array}$

Conclusion:

Therefore, it is proved that the drag force is $F_{\text{drag}}=810\text{N}$ .