A small steel ball falls through a syrup at a constant speed of 1.0m/s. If the steel ball is pulled upwards with a force equal to twice its effective weight, how fast will it move upward?

A small steel ball falls through a syrup at a constant speed of 1.0m/s

1.	A small steel ball falls through a syrup at a constant speed of 1.0m/s. If the steel ball is pulled upwards with a force equal to twice its effective weight, how fast will it move upward?
Answer» SOLUTION :`W_(e)=` effective WEIGHT In equilibrium `kv=w_(e)` Again `2w_(e)-w_(e)=kv^(1)` From eqs Iand II `v^(1)=v=1.0m//s`