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A 173 g object is attached to the end of an unstressed vertical spring (of constant 32.8 N/m) and then dropped. Acceleration of gravity is 9.8 m/s^2.

1. How far does it drop before coming to rest momentarily (in units of m)?

2. What is the maximum speed (in units of m/s)?

2006-10-16 06:32:58 · 2 answers · asked by Mariska 2 in Science & Mathematics Physics

2 answers

I think you can solve this from the energy equations. The energy of the spring at maximum extension will equal the the gravitational potential over the same distance.

For the maximum speed, that occurs where the spring force balances the gravitational force, as beyond this point, the mass decelerates. From the total energy calculated in the first part, subtract the gravitational potential and the spring energy at that point. What's left is the kinetic energy, from which you can get the speed.

2006-10-16 07:25:47 · answer #1 · answered by injanier 7 · 0 0

Yikes! No similar problems in the text!
I guess you'll have to apply what you know to solve an unknown problem rather than plug and chug. Synthesis and applied knowledge are great things.

Here's some hints: Hooke's law. The definition of a derivative of a function and how to find local max and min. All the forces balance or the mass will accelerate.

2006-10-16 14:00:35 · answer #2 · answered by Anonymous · 0 0

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