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
[tex]Ec = \frac{1}{2} * m * v^2[/tex]
Elastic potential energy is:
[tex]U = \frac{1}{2} * k * \Delta x^2[/tex]
The energy gained from the 0.2m pull will be turned into kinetic energy
Ec = U
Therefore:
[tex]\frac{1}{2} * m * v^2 = \frac{1}{2} * k * \Delta x^2[/tex]
Rearranging:
[tex]v^2 = \frac{k}{m} * \Delta x^2[/tex]
[tex]v = \Delta x * \sqrt{\frac{k}{m}}[/tex]
[tex]v = 0.2 * \sqrt{\frac{4.9}{0.1}} = 1.4 m/s[/tex]
After being released the ball will oscillate at the natural frequency of the system, which is
[tex]f = \frac{1}{2 * \pi} * \sqrt{\frac{k}{m}}[/tex]
And the period will be:
[tex]T = 2 * \pi * \sqrt{\frac{m}{k}}[/tex]
The period in this case is:
[tex]T = 2 * \pi * \sqrt{\frac{0.1}{4.9}} = 0.9 s[/tex]
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