Question

A box of mass 69g is attached to a horizontal spring with spring constant 123n/m. the spring is then extended by 11 cm, and the box is subsequently released. calculate the speed of the box, assuming the entire elastic potential energy is converted to the box's kinetic energy. give your answer to 2 decimal places if needed.

Answer 1

By applying the conservation of mechanical energy, the box released from the extended spring would have a speed of 4.64 m/s, with the entire elastic potential energy of the spring being converted into the box's kinetic energy.

Step-by-step explanation:

To calculate the speed of the box after it is released from the extended spring, we use the conservation of mechanical energy. The potential energy stored in the spring when it is extended is converted into the kinetic energy of the box.

The elastic potential energy (EPE) of the stretched spring is given by the formula:

EPE = ½ k x²

Where k is the spring constant and x is the extension of the spring. In this case, k = 123 N/m and x = 11 cm = 0.11 m.

EPE = ½ * 123 N/m * (0.11 m)²

EPE = ½ * 123 * 0.0121

EPE = 0.74315 J

Now, assuming this potential energy is fully converted into kinetic energy (KE) of the box, we use the KE formula:

KE = ½ m v²

Here, m is the mass of the box, and v is its velocity. Substituting the mass of the box, m = 69 g = 0.069 kg.

0.74315 J = ½ * 0.069 kg * v²

Solving for v, we get:

v² = (2 * 0.74315 J) / 0.069 kg

v² = 21.55862

v = √21.55862

v = 4.64 m/s

Therefore, the speed of the box is 4.64 m/s when it is released and the entire elastic potential energy is converted into kinetic energy.