How Is Kinetic Energy Calculated for a Raindrop Falling at Steady Speed?

[SUALEH MUSLIM]

Homework Statement

q)a rain drop of mass m is falling vertically through air with steady speed v the rain drop experiences retarding force kv,,k is constant,,acceleration of free fall is g,,k.e of drop is
a)mg/k
b)m.g^2/2k^2
c)m^3.g^2/k^2
d)m^3.g^2/2k^2
e)m^2.g/k

Homework Equations

f=kv...k.e=0.5mv^2

The Attempt at a Solution

guys i ve solved by imagining that drop is in dynamic equilibrium
f=mg
kv=mg
v=mg/k
so opt d,,,,but since in question it is not written whether drop is in dynaimic equilibrium or not so i am confused

 

[nrqed]

Homework Statement

q)a rain drop of mass m is falling vertically through air with steady speed v the rain drop experiences retarding force kv,,k is constant,,acceleration of free fall is g,,k.e of drop is
a)mg/k
b)m.g^2/2k^2
c)m^3.g^2/k^2
d)m^3.g^2/2k^2
e)m^2.g/k

Homework Equations

f=kv...k.e=0.5mv^2

The Attempt at a Solution

guys i ve solved by imagining that drop is in dynamic equilibrium
f=mg
kv=mg
v=mg/k
so opt d,,,,but since in question it is not written whether drop is in dynaimic equilibrium or not so i am confused

The question is stating that it is falling with a steady speed. This means that the speed is constant so yes, the net force is zero and your approach is correct.

 

[SUALEH MUSLIM]

but after that it is written it is acted upon by force kv,,so it means that v would have changed ,,and it is not in eq for time being

 

[AlephZero]

No, it means the retarding force = kv for any value of v. In the question there is just one constant value of v, which you found correctly as v = mg/k.

 

[rude man]

It's a steady-state problem. When the drop begins to fall, v builds up as mg - kv = ma, a = acceleration, but then levels out where there is no further acceleration so mg - kv = 0.