Energy and Simple Harmonic Motion.


by sailordragonball
Tags: energy, harmonic, motion, simple
sailordragonball
sailordragonball is offline
#1
Nov13-06, 12:48 AM
P: 42
A vertical spring with a spring constant of 470 N/m is mounted on the floor. From directly above the spring, which is unstrained, a 0.30 kg block is dropped from rest. It collides with and sticks to the spring, which is compressed by 2.5 cm in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in cm) above the compressed spring was the block dropped?

... I tried using KEo + PEo + SPEo = KEf + PEf + SPEf ... but, I can't get anywhere ...

... I found out that the frequency is 18.05hz and it's relative time is .056 seconds - and that's where I'm stuck!

HELP!
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sailordragonball
sailordragonball is offline
#2
Nov13-06, 12:55 AM
P: 42
I made a mistake ...

... I found a formula ...

... initial height = (2*mass*gravity)/(spring constant) ...

... but, there's the 2.5cm differential?

... what do I do?

Any help is gladly appreciated.
physics girl phd
physics girl phd is offline
#3
Nov13-06, 03:02 AM
physics girl phd's Avatar
P: 938
I'd use conservation of energy like you originally suggested. the 2.5 cm compression of the spring comes in there.

sailordragonball
sailordragonball is offline
#4
Nov13-06, 06:14 PM
P: 42

Energy and Simple Harmonic Motion.


I tried that ... but, the lingering the final displacement of 2.5cm is messing me up ... I was wondering ...

(mass*gravity)(initial height + 2.5cm) = .5(spring constant)[(2.5)^2] ...

... your thoughts?
physics girl phd
physics girl phd is offline
#5
Nov13-06, 06:29 PM
physics girl phd's Avatar
P: 938
First -- decide where the potential energy is defined as ZERO. You can put this anywhere... but be consistent.

Then -- look not just at the point where the brick is released, and the point where the brick is compressing the spring, but also think about the intermediate point (when the brick first starts to compress the spring).
Think about that types of energy (PE and or KE) are at each location.

Finally -- make sure the total energy at each location (PE+KE) is equal to the total energy at each other location (PE+KE).
sailordragonball
sailordragonball is offline
#6
Nov14-06, 02:33 PM
P: 42
I figured it out:

k = 470(N/m) - given
m = .3kg - given
x = 2.5 cm ... but use 0.025m in the arithmetic
g = 9.8 - given
h = initial height

(mass*gravity)(initial height + displacement) = .5(spring constant)[(displacement)^2]

(.3*9.8)(initial height + .025) = .5(470)[(.025)^2]

2.94(initial height + .025) = .146875

2.94(initial height) + .0735 = .146875

[initial height = .0249m] or [initial height = 2.49cm]

Use initial height = 2.49cm for the remainder of the problem.


... now find total height ...

... total height = initial height + displacement ...

... total height = 2.49cm + 2.5cm ...

... total height = 4.99cm


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