Question

A 0.100 kg ball hangs from a spring of negligible mass. When the ball is hung on the spring and is at rest, the spring is stretched by 0.200 m from its equilibrium to a new equilibrium, in which the tension in the spring balances the weight of the ball. a) What is the spring constant of this spring?
b) The ball is then pulled down another 0.200 m and released from rest. What is the speed of the ball when it gets back to the new equilibrium position?
c) How long after release does it take for the ball to get back to the new equilibrium position (for the first time)?

Answer 1

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

$Ec = \frac{1}{2} * m * v^2$

Elastic potential energy is:

$U = \frac{1}{2} * k * \Delta x^2$

The energy gained from the 0.2m pull will be turned into kinetic energy

Ec = U

Therefore:

$\frac{1}{2} * m * v^2 = \frac{1}{2} * k * \Delta x^2$

Rearranging:

$v^2 = \frac{k}{m} * \Delta x^2$

$v = \Delta x * \sqrt{\frac{k}{m}}$

$v = 0.2 * \sqrt{\frac{4.9}{0.1}} = 1.4 m/s$

After being released the ball will oscillate at the natural frequency of the system, which is

$f = \frac{1}{2 * \pi} * \sqrt{\frac{k}{m}}$

And the period will be:

$T = 2 * \pi * \sqrt{\frac{m}{k}}$

The period in this case is:

$T = 2 * \pi * \sqrt{\frac{0.1}{4.9}} = 0.9 s$

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