Solve HW Question: Find Maximum Distance Frame Moves Down from Initial Position

[lala56]

HW question, help, momentum

"A 0.150 kg frame, when suspended from a coil spring, stretches the spring 0.05 m. A 0.2 kg lump of putty is dropped from rest onto the frame from a height of 30.0 cm (Fig. 8.38). Find the maximum distance the frame moves downward from its initial position."

This problem has been posted before, but I have been unable to use it to solve my question

so, here is what I tried:
the k for the spring = 29.4 using F=kx

then
i found the velocity of the putty as soon as it touched the frame
mgh=.5mv^2
v=2.42
now, find the velocity of the system (putty and the frame)

Mp=mass of putty
Vp=velocity of putty
Mf=mass of frame
MpVp=(Mp+Mf)V2
plug the v found above into Vp, and then V2=1.39

so, now use this:
K1+U1+Ue1=K2+U2+Ue2

K1=kinetic energy
U1= potential energy (gravity)Ue1 = potential energy (spring)

i get:
.5(.2+.15)(1.39)^2 + .5(29.4)(.05)^2 + mgh = .5(29.4)x^2 + mgx

this is how I believe it should be done, except I don't know what to put in place of the "mgh". What is the original potential energy before the spring stretches.

I basically need to find the "x".

thanks, any help is greatly appreciated

 

[Doc Al]

I recommend that you use the unstretched position as your zero point for calculating gravitational PE as well as spring PE. Using that, what would the initial height be?

 

[lala56]

hm, but there is no "unstretched" postition cause when the frame is hanging on it, it is stretched .05m, so it is streteched form the beginning, before the putty is dropped

 

[Doc Al]

Sure there's an unstretched position: that's what you are measuring spring PE with respect to. I see three positions: unstretched (h=0), "initial" (h= -0.05m), and lowest point (h = x).

 

[lala56]

yes, I said that initial height is h=.05, and plugged that into the
.5(.2+.15)(1.39)^2 + .5(29.4)(.05)^2 + mgh = .5(29.4)x^2 + mgx

but I plugged it into the potential energy for spring part of the above equation: 5(29.4)(.05)^2

is everything up there correct thus far? (i mean the reasoning)

Maybe I am not quite understanding what you are trying to say, is it that the initial height of the spring ALSO applies to the gravitational potential energy, so the "mgh" above would be mg(.05)?

i tried that, and the answer is incorrect

(or maybe I didnt quite get what you were trying to say)

 

[Doc Al]

yes, I said that initial height is h=.05, and plugged that into the
.5(.2+.15)(1.39)^2 + .5(29.4)(.05)^2 + mgh = .5(29.4)x^2 + mgx

but I plugged it into the potential energy for spring part of the above equation: 5(29.4)(.05)^2

is everything up there correct thus far? (i mean the reasoning)

Note that your final gravitational PE is mgx, using the same "x" used in calculating the spring PE. To me, that means you've chosen the unstretched position as your zero point for calculating gravitational PE. Did you mean to do that?

Maybe I am not quite understanding what you are trying to say, is it that the initial height of the spring ALSO applies to the gravitational potential energy, so the "mgh" above would be mg(.05)?

i tried that, and the answer is incorrect

Note that the initial height is 0.05 m below the unstretched position.

You can pick any point you want as your reference level for measuring gravitational PE, since all that matters is the change in PE. Just be consistent.

 

[lala56]

OOOOOOHHHHHHHH

YESSSSSS
ic what you'rse saying
i messed up
it should really be (x+.05)
and then I solve for x

what is the initial gravitational potential here?
.5(.2+.15)(1.39)^2 + .5(29.4)(.05)^2 + mgh = .5(29.4)(X+.05)^2 + mg(X+.05)

it will probably be mg(.05) because .05 is the initial stretch and this even applies to the gravitaitonal force

edit:
it wouldn't really matter though if i made it (x+.05) cause when its is just x, i can subtract .05 from the final answer I get for x

 

[Doc Al]

what is the initial gravitational potential here?
.5(.2+.15)(1.39)^2 + .5(29.4)(.05)^2 + mgh = .5(29.4)(X+.05)^2 + mg(X+.05)

First tell me what the final gravitational PE is? You have now changed the meaning of X to be "the additional amount of stretch beyond the initial 0.05 stretch". What are you using as your reference height for gravitational PE? Pick any point, but pick!

it will probably be mg(.05) because .05 is the initial stretch and this even applies to the gravitaitonal force

Not sure what you are doing here. If GPE is mg(.05), then you are saying that the initial position is 0.05 m above some point?

Try this: Measure GPE from the lowest point that the system reaches, which is the position of maximum stretch. Set GPE = 0 for that point. Given that, what's the initial GPE? Hint: What's the initial height measured from the lowest point?