Answer to Question #206079 in Physics for Mya

Question

A teen is trying to practice her aiming skills with her slingshot. It contains a stone that is pulled back in a rubber band. It takes a force of 50 N to stretch the band a full distance of 0.25m. What is the potential energy stored in the band when a 0.5g stone is loaded on the band?

When that is solved the next question is, with what speed does it leave the slingshot if all the potential energy is transferred into kinetic energy at the moment the rock is let go?

Answer 1

The spring constant of the band can be found from the Hook's law:

$$k = \dfrac{F}{x} = \dfrac{50N}{0.25m} = 200\dfrac{N}{m}$$

The potential energy then:

$$W = \dfrac{kx^2}{2} = \dfrac{200N/m\cdot (0.25m)^2}{2} = 6.25J$$

The kinetic energy of the stone is equal to its potential energy:

$$\dfrac{mv^2}{2} = \dfrac{kx^2}{2} = W$$

where $m = 0.5g = 5\times 10^{-4}kg$ is the mass of the stone, and $v$ is its speed. Thus, obtain:

$$v = \sqrt{\dfrac{2W}{m}} = \sqrt{\dfrac{2\cdot 6.25J}{5\times 10^{-4}kg}} \approx 158m/s$$

ANswer. 6.25J, 158m/s.