Question #270721

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

 

.


1
Expert's answer
2021-11-24T11:58:58-0500

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


PEs=KE,PE_s=KE,12kx2=12mv2,\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2,v=kx2m,v=\sqrt{\dfrac{kx^2}{m}},v=15 Nm×(5×102 m)23×104 kg=11.0 ms.v=\sqrt{\dfrac{15\ \dfrac{N}{m}\times(5\times10^{-2}\ m)^2}{3\times10^{-4}\ kg}}=11.0\ \dfrac{m}{s}.

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