Answer to Question #270721 in Classical Mechanics for kyungsoo

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 =

v= _____  m.s

−1

 

.

Answer 1

We can find the velocity of the rubber band at the launch from the law of conservation of energy:

$$PE_s=KE,$$ $$\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2,$$ $$v=\sqrt{\dfrac{kx^2}{m}},$$ $$v=\sqrt{\dfrac{15\ \dfrac{N}{m}\times(5\times10^{-2}\ m)^2}{3\times10^{-4}\ kg}}=11.0\ \dfrac{m}{s}.$$