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

a) The spring constant of the spring is 4.9 N/m. b) The speed of the ball when it gets back to the new equilibrium position is 1.4 m/s. c) It takes 0.90 seconds for the ball to get back to the new equilibrium position (for the first time).

Step-by-step explanation:

a) What is the spring constant of this spring?

To find the spring constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. Mathematically, it can be written as F = kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the spring. In this case, the weight of the ball (mg) is balanced by the tension in the spring. So, we have mg = kx. Rearranging the equation gives us k = mg/x. Plugging in the given values, we have k = (0.100 kg)(9.8 m/s^2)/(0.200 m) = 4.9 N/m.

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?

When the ball is released, it will undergo simple harmonic motion. The amplitude of the motion is the distance the ball is pulled down, which is 0.200 m. Using the equation v = sqrt(k/m)x, where v is the speed of the ball, k is the spring constant, m is the mass of the ball, and x is the displacement of the ball, we can calculate the speed. At the equilibrium position, the displacement is 0, so v = sqrt(4.9 N/m / 0.100 kg)(0.200 m) = 1.4 m/s.

c) How long after release does it take for the ball to get back to the new equilibrium position (for the first time)?

The time period of the ball's motion can be calculated using the formula T = 2pi sqrt(m/k), where T is the time period, m is the mass of the ball, and k is the spring constant. Plugging in the given values, we have T = 2pi sqrt(0.100 kg / 4.9 N/m) = 0.90 s.

Answer 2

a) 4.9 N/m

b) 1.4 m/s

c) 0.225 s

Step-by-step 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 = (1)/(2) * m * v^2`

Elastic potential energy is:

`U = (1)/(2) * k * \Delta x^2`

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

Ec = U

Therefore:

`(1)/(2) * m * v^2 = (1)/(2) * k * \Delta x^2`

Rearranging:

`v^2 = (k)/(m) * \Delta x^2`

`v = \Delta x * \sqrt{(k)/(m)}`

`v = 0.2 * \sqrt{(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 = (1)/(2 * \pi) * \sqrt{(k)/(m)}`

And the period will be:

`T = 2 * \pi * \sqrt{(m)/(k)}`

The period in this case is:

`T = 2 * \pi * \sqrt{(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