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 upwards?

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 upwards?
Answer» Solution :`W_(e)` = EFFECTIVE weight In equilibrium, `kv=w_(e)` Again, `2w_(e)-w_(e)=kv^(1)` From Eqs. (i) and (II) `v^(1)=v=1.0m//s`