(a) Apply energy conservation from start to top of spring using h = 0 as top of spring.
U = K
mgh = ½mv2
(9.8)(0.45) = ½v2
v = 3m/s
(b) At equilibrium the forces are balances
Fnet = 0
Fsp = mg =(2)(9.8) = 19.6 N
(c) Using the force from part b,
Fsp = k∆x
19.6 = 200∆x
∆x = 0.098m
(d) Apply energy conservation using the equilibrium position as h = 0.
(Note that the height at the start position is now increased by the amount of ∆x found in part c hnew = h + ∆x = 0.45 + 0.098 = 0.548m)
Utop = Usp + K
mgh = ½k∆x2 + ½ mv2
(2)(9.8)(0.548) = ½ (200)(0.098)2 + ½ (2)(v2)
v = 3.13m/s
(e) This is the maximum speed because this was the point when the spring force and weight were equal to each other and the acceleration was zero. Past this point, the spring force will increase above the value of gravity causing an upwards acceleration which will slow the box down until it reaches its maximum compression and stops momentarily.