Elastic potential energy

Question

A dart gun has a spring with

k= 222 N/m. A 0.0100 kg dart is

loaded, and the spring is pulled

back 0.0480 m. At what speed

does the dart leave the gun?

Unit=m/s

if someone could please Thoroughly work out this problem I would greatly appreciate it ,I would like to compare my process .

NO EXTIOR FACTORS NEEDED TO BE TAKIN INTO ACCOUNT …Thank you so much:))

Answer 1

The potential energy of the spring (U_S) is calculated from:

U_S = 1/2kx² where k is the spring constant and x is the distance the spring is pulled back.

After the dart is released, all that energy is converted to kinetic energy as the dart leaves the spring/gun. Kinetic energy is calculated as 1/2mv². Since energy is conserved, the spring potential energy equals the kinetic energy so:

1/2kx² = 1/2mv²

You can just plug in the numbers and solve for "v".