Question

A 0.20 kg mass is hung from a vertical spring of force constant 55 N/m. When the spring is released from its unstretched equilibrium position, the mass is allowed to fall. Use the law of conservation of energy to determine (a) the speed of the mass after it falls 1.5 cm

Answer 1

The speed of the mass after it falls 1.5 cm can be determined using the law of conservation of energy. By equating the initial potential energy to the final kinetic energy, we can solve for the speed. The speed is approximately 0.542 m/s.

Step-by-step explanation:

To determine the speed of the mass after it falls 1.5 cm, we can apply the law of conservation of energy. Initially, the mass has potential energy due to the gravitational force (mgh) and no kinetic energy. As it falls, this potential energy is converted into kinetic energy. Using the equation for potential energy (PE = mgh) and the equation for kinetic energy (KE = 1/2mv^2), we can equate the initial potential energy to the final kinetic energy to solve for the speed (v).

Given:

• Mass (m) = 0.20 kg
• Force constant (k) = 55 N/m
• Height (h) = 1.5 cm = 0.015 m

Using the equation PE = KE:

mgh = 1/2mv^2

Simplifying:

0.20 kg * 9.8 m/s^2 * 0.015 m = 1/2 * 0.20 kg * v^2

Solving for v:

v^2 = 0.015 m * 9.8 m/s^2 * 2

v^2 = 0.294 m^2/s^2

v ≈ 0.542 m/s

Therefore, the speed of the mass after it falls 1.5 cm is approximately 0.542 m/s.