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A novelty clock has a 0.0100 -kg mass object bouncing on a spring that has a force constant of 1.25 $\mathrm{N} / \mathrm{m} .$ What is the maximum velocity of the object if the object bounces 3.00 $\mathrm{cm}$ above and below its equilibrium position? (b) How many joules of kinetic energy does the object have at its maximum velocity?

$\begin{aligned} V_{\max } &=0.3354 \mathrm{m} / \mathrm{s} \\ E_{\max } &=5.625 \cdot 10^{-4} \mathrm{J} \end{aligned}$

Physics 101 Mechanics

Physics 103

Chapter 16

Oscillatory Motion and Waves

Periodic Motion

Wave Optics

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here we can equate the of ah maxim kinetic energy 1/2 be max squared, equaling to the potential energy of a spring. And so we're simply using for part of the conservation of mechanical energy. And so we can say that the maximum velocity isn't gonna be equaling X multiplied by the spring. Constant divided by the square root of the spring. Constant divided by the mass. And so this would be equal in point 300 meters multiplied by the square root of 1.25 newtons per meter. This would be divided by 0.100 kilograms and we find a maximum velocity equaling point 335 meters per second. This would be our maximum velocity. Four part, eh, four part B. They want a maximum potential energy and this would be equaling the potential energy of a spring at its amplitude or its maximum displacement. So this would be 1/2 K x squared, so this would be 1/2 times the spring constant, which we know to be 1.25 Newtons per meter and then this would be multiplied by its maximum displacement of 0.3 meters quantity squared. And so we find that our maximum potential energy would then be equaling 5.63 times 10 to the negative fourth. Jules, this would be our final answer for part B. That is the end of the solution. Thank you for one.

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