Work, Springs

  • #1

Homework Statement



A mass of 122 g is attached to a vertically hung spring. The mass stretches the spring 13.8 cm. (a) What is the spring constant of the spring? (b) If the mass is dropped from rest from a position where the spring is just relaxed, what will be the maximum distance the mass will stretch the spring?


Homework Equations


W = ΔK + ΔP
Gravitational P.E = mgh
Spring P.E = 1/2kx²


The Attempt at a Solution



(a) What is the spring constant of the spring?
I used a FBD to see that the only relevant force working here would be the force of gravity....so mg.

spring constant = Force / Δx

We're given Δx = .138m

Force would be just the mg I believe...so if that is true

K = F / Δx = 1.1956 / .138 = 8.66 for the spring constant


(b) If the mass is dropped from rest from a position where the spring is just relaxed, what will be the maximum distance the mass will stretch the spring?

For here I'm not too sure....I tried using Work = Δk + ΔP

so I did

mgh (also written as mgΔy) for the gravitational potential energy + 1/2Kx² for the spring potential energy, and since velocity would be 0 the kinetic energy wouldn't apply

so

mgΔy = 1/2Kx²

I'm not sure if this is the right way to be approaching this...

any hints here?
 

Answers and Replies

  • #2
Doc Al
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(a) What is the spring constant of the spring?
I used a FBD to see that the only relevant force working here would be the force of gravity....so mg.

spring constant = Force / Δx

We're given Δx = .138m

Force would be just the mg I believe...so if that is true

K = F / Δx = 1.1956 / .138 = 8.66 for the spring constant
Good. (Don't forget units.)

(b) If the mass is dropped from rest from a position where the spring is just relaxed, what will be the maximum distance the mass will stretch the spring?

For here I'm not too sure....I tried using Work = Δk + ΔP

so I did

mgh (also written as mgΔy) for the gravitational potential energy + 1/2Kx² for the spring potential energy, and since velocity would be 0 the kinetic energy wouldn't apply

so

mgΔy = 1/2Kx²
That's the right idea. Hint: Measure gravitational PE from the lowest point.
 
  • #3
Good. (Don't forget units.)
Ahh sorry....8.66N/m

That's the right idea. Hint: Measure gravitational PE from the lowest point.

well see that's where I can never seem to figure it out....from the lowest point...we don't know what that lowest point will be

mgΔy = 1/2Kx²

if it's at the very bottom...the gravitational potential energy will be 0 correct?
 
Last edited:
  • #4
Doc Al
Mentor
45,093
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Ahh sorry....8.66N/m
Good.


well see that's where I can never seem to figure it out....from the lowest point...we don't know what that lowest point will be
That's what you'll figure out.

mgΔy = 1/2Kx²

if it's at the very bottom...the gravitational potential energy will be 0 correct?
Yes.

Think in terms of energy conservation. At the top, the spring is unstretched and all the energy is gravitational PE. At the bottom, the spring is fully stretched and the gravitational PE = 0 (assuming you measure from that point).

You have the equation. But how are Δy and x related?
 
  • #5
Yes.

Think in terms of energy conservation. At the top, the spring is unstretched and all the energy is gravitational PE. At the bottom, the spring is fully stretched and the gravitational PE = 0 (assuming you measure from that point).

You have the equation. But how are Δy and x related?

I really need to fully comprehend energy conservation and how energy's are converted to each other....just haven't found that 1 source that makes it "click" in my mind.

but back to this

at very first thought I would want to say that the Δy and x will be the same number....

Δy would be the change in height...and x would be the amount the string is stretched....so yeah I'm going to say it's the same number?
 
  • #6
Doc Al
Mentor
45,093
1,398
at very first thought I would want to say that the Δy and x will be the same number....

Δy would be the change in height...and x would be the amount the string is stretched....so yeah I'm going to say it's the same number?
Exactly! So rewrite your equation and solve for x.
 
  • #7
Exactly! So rewrite your equation and solve for x.

mgΔy = 1/2Kx²
mg = 1/2Kx
x = 2mg / K

But the thing is, would I still be able to use that same spring constant as obtained in part (a)?

I keep thinking no because of the fact that wouldn't the (although small) acceleration that the mass picks up while falling cause the spring to stretch a bit farther than if the mass is just freely hanging from it? or would that value be so small as to be negligible?
 
  • #8
Doc Al
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45,093
1,398
mgΔy = 1/2Kx²
mg = 1/2Kx
x = 2mg / K
Good. Note that the total amount of stretch is twice what it was when the mass was just hanging there.

But the thing is, would I still be able to use that same spring constant as obtained in part (a)?
Of course. It's the same spring.

I keep thinking no because of the fact that wouldn't the (although small) acceleration that the mass picks up while falling cause the spring to stretch a bit farther than if the mass is just freely hanging from it? or would that value be so small as to be negligible?
The speed that the falling mass picks up as it falls allows it to stretch twice as much at its furthest point. That's far from negligible. (If that's what you meant.)

Note that the mass will oscillate about the equilibrium point. It starts a distance 13.8 cm above equilibrium and reaches 13.8 cm below equilibrium. And then continues to oscillate between those two points. This oscillation about the equilibrium point is a useful thing to remember.
 

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