Question

17 a 75 - kg skydiver jumps out of an airplane. before she opens her parachute, the force of air resistance on her body is 550 n. what is the net acceleration of the skydiver at that point?

• 2.47 m/s²

0 m/s²

• 185 n
• 9.8 m/s²

Explanation:

Step1: Calculate the gravitational force

The gravitational force \(F_g = mg\), where \(m = 75\space kg\) and \(g= 9.8\frac{m}{s^{2}}\). So \(F_g=75\times9.8 = 735\space N\) (downward).

Step2: Calculate the net force

The air - resistance \(F_{air}=550\space N\) (upward). The net force \(F_{net}=F_g - F_{air}\). So \(F_{net}=735 - 550=185\space N\) (downward).

Step3: Calculate the acceleration

Using Newton's second law \(F_{net}=ma\), then \(a=\frac{F_{net}}{m}\). Substitute \(F_{net} = 185\space N\) and \(m = 75\space kg\), we get \(a=\frac{185}{75}\approx2.47\frac{m}{s^{2}}\) (downward, so the value is positive if we take down as positive direction, but if we consider the options, maybe there is a sign - convention issue. If we take up as positive, \(a=\frac{F_{air}-F_g}{m}=\frac{550 - 735}{75}\approx - 2.47\frac{m}{s^{2}}\))

Answer:

• \(2.47\frac{m}{s^{2}}\)