(a) 1.08 J
The elastic potential energy stored in the block at any position x is given by
where
k is the spring constant
x is the displacement relative to the equilibrium position
Here we have
k = 860 N/m
x = 5.00 cm = 0.05 m is the position of the block
Substituting, we find
(b) 1.16 m/s
The total mechanical energy of the spring-mass system is equal to the potential energy found at point (a), because there the system was at its maximum displacement, where the kinetic energy (because the speed is zero).
At the equilibrium position, the mechanical energy is sum of kinetic and potential energy
E = K + U
However, at equilibrium position x = 0, so U = 0. Therefore, the kinetic energy is equal to the total energy found at point (a)
where
m = 1.60 kg is the mass of the block
v is the speed
Solving for v, we find
(c) 1.00 m/s
When the block is at position x = 2.50 cm, the mechanical energy is sum of kinetic and potential energy:
where
E = 1.08 J is the total mechanical energy
m = 1.60 kg is the mass
v is the speed
k = 860 N/m
x = 2.50 cm = 0.025 m is the displacement
Solving for v, we find