A spring (k = 400 N/m) is hung vertically.
a) If a 5 kg mass is attached to the end of the spring and gently lowered to rest position, what will be the stretch of the spring?
b) If the 5 kg mass is attached and simply dropped, what will be the maximum velocity and the maximum stretch?
I know the anwer to the first part. F=- kx so solving for x = .1225 m
I am looking for the answer to part B.
(Please also refer to the attachment)
Initially the spring is un-stretched with mass attached to it and held by hand which exerts an upward force on the mass to prevent it accelerating down under gravity. As the mass is gradually lowered (all the time held with hand), the spring gets stretched generating an upward restoring force. The upward force applied by hand must be reduced by the same magnitude such that at any instant net upward force (restoring force in the spring + force applied by hand) is equal to the weight (downward). As the mass continues to move down, a stage comes when the restoring force due to the spring is equal to the weight and the force applied by hand is zero (i.e. the mass can be let go). The mass stays in equilibrium in this position as net force on it is zero. ...
Detailed step by step solution provided.