# A 66.9 g mass is attached to the end of an unstressed vertical spring

• Aerospace
In summary, to find the maximum speed of a 66.9 g mass attached to an unstressed vertical spring, you can use the conservation of energy principle and Hooke's law to calculate the kinetic, gravitational potential, and elastic potential energies at the initial and final states. You can also use the fact that the object will be moving fastest at the equilibrium position.

#### Aerospace

A 66.9 g mass is attached to the end of an unstressed vertical spring (of constant 63 N/m) and then dropped. What is its maximum speed? How far does it drop before coming to rest momentarily?

If anyone could help?

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Aside from knowing some clever formulae for the displacement, velocity, and acceleration of objects attached to springs (which you may know at this point!) the best approach is conservation of energy.

You know the kinetic energy (KE), gravitational potential energy (GPE), and elastic potential energy (EPE) of the system at the initial time.

You know the KE at the final time, so conservation of energy allows you to write an equation in the two unknowns, final GPE and final EPE. Applying Hooke's law gives you a second equation in these unknowns, so you can find the GPE and EPE for the final state, and thus find displacement.

You should also know that an object attached to a spring is moving fastest at the equilibrium position which is half-way between the two extreme positions. You should now be able to compute the GPE and EPE at this point, allowing you to solve for KE by using conservation of energy, and thus giving you velocity.

(If you don't know this fact, then you can apply conservation of energy to get a formula for the velocity at any point, and then use techniques of algebra or calculus to find the maximum)