Question #160921
A 50.0 g object connected to a spring with a force constant of 35.0N/m oscillates with amplitude of 4.00cm on a frictionless, horizontal surface. Find
(a) the total energy of the system
(b) the speed of the object when its position is 1.00cm
1
Expert's answer
2021-02-12T16:17:51-0500

(a) The total energy of the system can be found as follows:


E=12kA2,E=\dfrac{1}{2}kA^2,E=1235.0 Nm(0.04 m)2=28103 J.E=\dfrac{1}{2}\cdot35.0\ \dfrac{N}{m}\cdot(0.04\ m)^2=28\cdot10^{-3}\ J.

(b) The speed of the object can be found from the formula:


v=km(A2x2),v=\sqrt{\dfrac{k}{m}(A^2-x^2)},v=35.0 Nm0.05 kg((0.04 m)2(0.01 m)2)=1.02 ms.v=\sqrt{\dfrac{35.0\ \dfrac{N}{m}}{0.05\ kg}((0.04\ m)^2-(0.01\ m)^2)}=1.02\ \dfrac{m}{s}.

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