Answer:
a) 4.9 N/m
b) 1.4 m/s
c) 0.225 s
Explanation:
Hooke's law states that
F = k * Δx
Where
F: force applied to a spring
k: constant of the spring
Δx: elongation of the spring
The force applied in this case is the weight of the ball, this is
P = m * g = 0.1 kg * 9.81 m/s^2 = 0.981 N
Rearrainging Hooke's law:
k = F / Δx
k = 0.981 / 0.2 = 4.9 N/m
If the ball is pulled down the spring will acquire some potential energy, when it is released, the potential energy will be released as kinetic energy on the ball
Elastic potential energy is:
The energy gained from the 0.2m pull will be turned into kinetic energy
Ec = U
Therefore:
Rearranging:
After being released the ball will oscillate at the natural frequency of the system, which is
And the period will be:
The period in this case is:
The ball will move up and down taking T time to complete a cycle, the movement from the stretched position to the equilibrium position takes T/4 = 0.225 s