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

[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?

I wasn't sure how to go about this problem.

If anyone could help?

 

[Hurkyl]

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)

 

[M98Ranger]

The maximum speed of the mass can be calculated using the equation for potential energy and kinetic energy:

PE = mgh

KE = 1/2mv^2

At the top of the drop, all of the potential energy is converted into kinetic energy, so we can set these two equations equal to each other:

mgh = 1/2mv^2

We know the mass (66.9 g) and the gravitational acceleration (9.8 m/s^2), so we can solve for v:

v = √(2gh)

Plugging in the values, we get:

v = √(2 * 9.8 m/s^2 * 0.0669 kg * 0.63 N/m) = 0.76 m/s

So the maximum speed of the mass is 0.76 m/s.

To find the distance the mass drops before coming to rest momentarily, we can use the equation for potential energy again:

PE = mgh

At the bottom of the drop, all of the kinetic energy is converted back into potential energy, so we can set these two equations equal to each other:

1/2mv^2 = mgh

Solving for h, we get:

h = v^2/2g

Plugging in the values, we get:

h = (0.76 m/s)^2/(2 * 9.8 m/s^2) = 0.0308 m = 3.08 cm

So the mass drops approximately 3.08 cm before coming to rest momentarily.