Confused about springs

My physics class is intro more or less (it's calc based; 111 NOT 101). I'm a bit confused when it comes to springs..specifically, finding the maximum compression a spring will have.

Here are the given variables in a problem in my textbook:

*spring has negligible mass
*force constant(k) = 1600n/m
*spring is placed virtically with one end on the floor
*1.2kg book is dropped from a height of .80m above the top of the spring

maximum distance the spring will be compressed?

what confuses me specifically are the signs (+/-) and the distances involved. Here is what I am thinking:

I need to use K1 + U1 = K2 + U2 (k1 = zero)

now when it comes to potential energy, i don't know whether to use MGY or 1/2kx^2..can someone point me in the right direction? thanks!
 

Doc Al

Mentor
44,642
966
NutriGrainKiller said:
Here is what I am thinking:
I need to use K1 + U1 = K2 + U2 (k1 = zero)
Good. (What will K2 be when the the spring is maximally compressed?)
now when it comes to potential energy, i don't know whether to use MGY or 1/2kx^2.
Since the height of the book changes, you need gravitational PE; since the spring is compressed, you need spring PE. Use both!
 
Doc Al said:
Good. (What will K2 be when the the spring is maximally compressed?)
So are you saying that K2=zero as well? i guess that would make sense

Doc Al said:
Since the height of the book changes, you need gravitational PE; since the spring is compressed, you need spring PE. Use both!
So..I need U1 (spring PE) + U2 (grav PE) = U1 (spring PE) + U2 (grav PE)

-or-

1/2MV1^2 + mgy1 = 1/2MV2^2 + mgy2 ?

EDIT: Another question; I'm having trouble differentiating between the Y's..where should I solve for Y and where should I plug in .8?
 
Last edited:

Doc Al

Mentor
44,642
966
NutriGrainKiller said:
So are you saying that K2=zero as well? i guess that would make sense
If K2 wasn't zero, the spring could not be fully compressed--the book would still be moving.
So..I need U1 (spring PE) + U2 (grav PE) = U1 (spring PE) + U2 (grav PE)
That will work.
-or-
1/2MV1^2 + mgy1 = 1/2MV2^2 + mgy2 ?
This won't work--you left out spring PE.
EDIT: Another question; I'm having trouble differentiating between the Y's..where should I solve for Y and where should I plug in .8?
Call the amount that the spring compresses "x". Then pick a reference point for gravitational PE (may as well call the unstretched position of the spring to be y = 0) and set up the equation. Give it a shot.
 
Doc Al said:
This won't work--you left out spring PE.
sorry - should have 1/2KX1^2 + MGY1 = 1/2KX2^2 + MGY2

x is how much the spring is compressed. 0 is where the spring is at rest (so -x would be the compression)

1/2(1600)x^2 + (1.2)(9.8)(.8) = 1/2(1600)x^2 + (1.2)(9.8)(.8)

....it's the same on both sides :confused: would cancel out..
 

Doc Al

Mentor
44,642
966
Initially the spring is uncompressed, so x1 = 0. Initially the book is at y1 = 0.8; but where is it when the spring is compressed by amount "x"?
 
Doc Al said:
Initially the spring is uncompressed, so x1 = 0. Initially the book is at y1 = 0.8; but where is it when the spring is compressed by amount "x"?

...-x?

so here's what i have now:

mgy1 = 1/2kx2^2 + mg(-x)

or

9.408 = 800x2^2 - 11.76x <--looks like quadratic
 

Doc Al

Mentor
44,642
966
Good! Now you're cooking. Solve that equation.
 
somehow i usually tend to screw the quadratic equation up, but here is what i got:

(11.76 +/- sqrt(30243.9))/1600

which ends up being: .116m or 11.6cm

the answer in the book was 12cm, i don't think the sig figs required that type of rounding
 

Doc Al

Mentor
44,642
966
One of those solutions should be negative. That one can be discarded as physically irrelevant.

(Your initial data was only good to two sig figs; so your answer should be rounded off accordingly.)
 
Doc Al said:
One of those solutions should be negative. That one can be discarded as physically irrelevant.
so how do i get the answer if this equation is discarded?
 

Doc Al

Mentor
44,642
966
Don't discard the equation; discard one of the solutions.

As often happens with quadratic equations for physical systems, only one of the solutions makes any physical sense. In this case, a negative answer doesn't make sense so you know that the real answer is the other one.
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top