Question

A box of mass 37g is attached to a horizontal spring with spring constant 106N/m. The spring is then extended by 9 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

The speed of the box attached to the spring is found to be 4.77 m/s after converting the elastic potential energy to kinetic energy.

Step-by-step explanation:

The student asked how to calculate the speed of a box attached to a spring when its entire elastic potential energy is converted to kinetic energy. The mass of the box is 37g (which we will convert to kilograms) and the spring constant is given as 106 N/m. The box is extended by 9 cm from the equilibrium position and released.

To solve for the speed, we first find the elastic potential energy using the formula:

• U = 1/2 kx²

where U is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position. We plug in the values:

U = 1/2 x 106 N/m x (0.09 m)² = 0.4284 Joules

Assuming all potential energy is converted into kinetic energy:

• KE = U

where KE is the kinetic energy.

Next, we use the kinetic energy formula to find the speed:

• KE = 1/2 mv²

0.4284 J = 1/2 x 0.037 kg x v²

Solving for v, the speed of the box, we get:

v = sqrt((2 x 0.4284 J) / 0.037 kg) which yields v = 4.77 m/s when rounded to two decimal places.