Answer to Question #106052 in Classical Mechanics for Jimar Che

Question

A rubber band has mass m = 0.30 g and a spring constant k = 15 N.m^−1 . I stretch it by 5.0 cm (which in this case doubles its length). Assume the rubber band behaves as a Hooke's law spring. Assume that, when you launch the rubber band, all of the stored potential energy is converted into kinetic energy. How fast is it at the launch?

v = _____ m.s^−1 . ( USE THE CORRECT SIGNIFICANT NUMBER FOR THE FINAL ANSWER)

Answer 1

We will apply energy conservation here i.e initial sum of kinetic and potential energy is equal to final sum of kinetic and potential energy

$K_i+U_i=K_f+U_f$

$0+\frac12kx^2=\frac12mv^2+0$

$15\times(0.05)^2=0.3\times10^{-3}\times v^2\\v^2=\frac{125}3m/s\\v=6.45m/s$