Question #173081

a slingshot has 12 J of elastic potential energy when it screeched  to 0.2 m away from its equilibrium. What is the spring constant for the slingshot




1
Expert's answer
2021-03-18T20:11:04-0400
PE=12kx2,PE=\dfrac{1}{2}kx^2,k=2PEx2=212 J(0.2 m)2=600 Nm.k=\dfrac{2PE}{x^2}=\dfrac{2\cdot12\ J}{(0.2\ m)^2}=600\ \dfrac{N}{m}.

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