Problem using Newtons second law

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A sky-diver experiences two forces: the downward gravitational force (Fg = mg) and the upward air resistance force (Fr), which is proportional to velocity (Fr = kv). Using Newton's Second Law, the net force can be expressed as Fnet = Fg - Fr. This leads to the equation mg - kv = ma, where the negative sign indicates that air resistance opposes the downward motion. The resulting acceleration can be formulated as a = g - (kv/m), clarifying the relationship between acceleration and velocity.
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Once her chute opens, a sky-diver of mass m is acted upon by a downward force Fg due to gravity, and an upward force Fr due to air resistance. If Fg = mg, where g is gravitational acceleration, and Fr is proportional to velocity v, use Newton’s Second Law of Motion to write acceleration a as a function of velocity v.

I don't know if I'm on the right track
Fnet = ma
fnet = Fg+ Fr
Fg = mg

Mg + Fr = ma
 
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yes on right track except the signs. Now since F_r is proportional to velocity v, we can write

F_r=kv where k is constant of proportionality.

\Rightarrow mg-kv=ma

the negative sign is there since the air drag opposes the downward motion,

\therefore a = g-\frac{kv}{m}
 
IssacNewton said:
yes on right track except the signs. Now since F_r is proportional to velocity v, we can write

F_r=kv where k is constant of proportionality.

\Rightarrow mg-kv=ma

the negative sign is there since the air drag opposes the downward motion,

\therefore a = g-\frac{kv}{m}

That makes a lot more sense, thank you.
 
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