Elastic potential energy

Question

A dart gun has a spring with

k= 185 N/m. A 0.0100 kg dart is

loaded, and the spring is pulled

back 0.0520 m. If the dart is shot

straight up, how high does it go?

Unit =m

IF SOMEONE COULD ANSWER THIS Thoroughly I want to check my answer ,I would so greatly appreciate it !!

Answer 1

The potential energy stored in the spring is calculated using:

U_S = 1/2kx² where "k" is the spring constant and "x" is the distance the spring is compressed so, plugging in the values for "k" and "x", we get:

U_S = 1/2(185)(0.0520)² = 0.25012 joules

IF you assume no energy losses due to air resistance or friction or anything else, then after the dart leaves the spring/gun, it will have the same amount of energy, all converted to kinetic energy (this is what conservation of energy means), then, when the dart has reached its maximum height, it will still have the same amount of energy, now all converted into gravitational potential energy (calculated as U_G = mgh).

So, we can say 0.25012 = mgh or 0.25012 = (0.0100)(9.81)(h). Then you can solve for "h"