I don\'t understand why \"net force= mg minus kv^2\". I know that \"mg=w\" (mass

Question

I don't understand why "net force= mg minus kv^2". I know that "mg=w" (mass times the acceleration of gravity equals weight), and also that "kv^2=the force of drag". But why does "net force=weight minus drag"?

The question in my (high school) physics book is "A 65.0-kg skydiver reaches a terminal speed of 55.0 m/s with her parachute undeployed. Suppose the drag force acting on her is proportional to the speed squared, or Fdrag = kv^2 (that is, 'k' times 'v squared'). (a) What is the constant of proportionality k? (Assume the gravitational force is 9.80 m/s^2.)" For the answer to (a) I got k=0.211 kg/m. I checked that in the back of the book, and it is correct. But here's the catch: I got (b) wrong. Take a look at what I did:

V=55.0/2=27.5 m/s
Fdrag=kv^2
I then substituted mg (mass times the acceleration of gravity) for Fdrag. So:
m=65.0 kg
k=0.211 kg/m
So:
65.0 x a = (0.211)(27.5)^2
65.0 x a = 159.56875
a= 2.454903846 m/s^2
with significant figures: 2.45 m/s^2
BUT...the answer in the back of the book is 7.35 m/s^2...huh???

Explanation / Answer

Forces acting are : (1) Weight of the body W = mg    (Downward) (2)Drag force F = Kv^2   (opposite to the motin of the body, i.e, Upward) For vector quantities we have to consider the direction, It is very important. Resultant force = Vector sum of the forces acting on the body                    ma = mg + (- F)                   ma = mg - F                  65*a = 637 - 159.56                   65a = 477.43                         a = 7.345 m/s^2   Henc eAcceleration a = 7.345 m/s^2                

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