Physics Conservation of Energy Problem Solving

Question

A marble launcher is set up to launch marbles in the horizontal direction. It has a spring constant of 190 N/m and a marble of mass 0.0059 kg is loaded and compresses the spring 0.071 m. What is the intitial speed of the marble when it exits the launcher?

Answer 1

Thanks to the Law of Conservation of Energy, we know that all of the potential energy stored in the compressed spring will be converted to kinetic energy in the marble. Thus, we can set the two quantities equal to each other, and solve for the initial speed:

PE_spring = kx²/2 (k = spring constant, x = displacement of spring from equilibrium)

KE_marble = mv²/2 (m = mass, v = speed)

kx²/2 = mv²/2; (190 N/m)(0.071 m)²/2 = (0.0059 kg)(v²)/2; v = 12.7 m/s

I hope this helped!