Ch 9.3 Help

Go down

Ch 9.3 Help

Post  Guest01 on Mon Oct 27, 2008 9:31 pm

Question 1
A compressed spring with a spring constant of 129.4 N/m has 8.015 J potential energy stored in it. How far is the spring compressed from its relaxed state (see sheet 22)? Indicate with a positive (negative) sign whether the spring has a different (the same) potential energy stored in it when it is stretched by the same amount.

How To Solve
Answer is



Question 2
A spring with a spring constant of 53.899 N/m which is aligned vertically is compressed by 18.426 cm from its relaxed state. A 0.1338 kg mass is then placed on the spring and the spring is released. What is the kinetic energy of the mass as it lifts off from the relaxed spring (see sheet 24)? Indicate with a negative (positive) sign whether the mass gains and the spring looses potential energy (or the other way around).

How To Solve
Answer is



Question 3
A 0.5295 kg block attached to a spring with a spring constant of 52.3 N/m oscillates frictionless with an amplitude of 0.2722 m on a horizontal plane. What is the maximum velocity of the block (see sheet 27)? Indicate with a negative (positive) sign whether one can calculate from the data given the total energy of the mass-spring system for any given time during the oscillation (or not).

How To Solve
Answer is



Question 4
A pendulum with a string length of 52.47 cm swings back and forth. You push it whenever it has come to rest at a turning point. With which frequency do you have to push it in order to cause resonance, that is maximal energy build up in the oscillation (see sheet 29,33,34)? Indicate with a negative (positive) sign whether your result depends (does not depend) on the mass suspended from the string.

How To Solve
Answer is



[Note: Currently working on it]

Guest01

Posts : 133
Join date : 2008-09-19

View user profile

Back to top Go down

ch 9.3 Questions

Post  hwilson on Tue Oct 28, 2008 1:57 pm

Could someone please help with these I have been working on them for a while and I am having a lot of trouble?

hwilson
Guest


Back to top Go down

Re: Ch 9.3 Help

Post  Meg on Tue Oct 28, 2008 3:37 pm

...I have too and I am really having trouble...help please

Meg
Guest


Back to top Go down

Ch 9.3 Help

Post  sedwards on Tue Oct 28, 2008 7:44 pm

I am so confused could someone please help with these questions?

sedwards
Guest


Back to top Go down

9.3 Quiz

Post  gguest on Tue Oct 28, 2008 8:12 pm

Help?

gguest
Guest


Back to top Go down

Re: Ch 9.3 Help

Post  Guest01 on Tue Oct 28, 2008 8:47 pm

Easy there guys, it's not due until another 4 days.

What do you not exactly get? What does your work look like?

Guest01

Posts : 133
Join date : 2008-09-19

View user profile

Back to top Go down

Re: Ch 9.3 Help

Post  gwar on Tue Oct 28, 2008 10:24 pm

Are you guys even trying... some of this is right off of the lecture slides directly. If you're actually having trouble, post up your work and I'll correct it but I mean... c'mon.

gwar
Guest


Back to top Go down

quiz questions

Post  guest22 on Tue Oct 28, 2008 10:31 pm

I have number 1 and number 3? Can anyone help with 4 I am still working on number 2?

guest22
Guest


Back to top Go down

Question 3

Post  angelbab on Tue Oct 28, 2008 11:18 pm

Does anyone have number 3?

angelbab
Guest


Back to top Go down

Answers

Post  helpme on Wed Oct 29, 2008 7:14 pm

1. negative
2. negative
3. negative
4. positive

1. Formula: [(2*PE)/K]^.5 = X
2. Formula: V^2= sqrt[(.5kx^2-mgx)/.5m]
Then KE= .5mv^2
3. Formula: Vo^2= [K*(x_0^2)]/m
Then, take the sqrt
4.Formula: T= 2pi*sqrt(L/g)
Then divide 1 by T to get f. So if T=2, f= 1/2= .5
Don't forget to change to meters. Hope this helps, good luck.

helpme
Guest


Back to top Go down

question 2

Post  101 on Wed Oct 29, 2008 11:05 pm

for the formula of question 2 there shouldn't be a sqrt since you're looking for v^2

101
Guest


Back to top Go down

Question 2

Post  Meg on Thu Oct 30, 2008 1:18 am

Can someone pleease help me?? I keep getting #2 wrong...

Here's what I did...

A spring with a spring constant of 56.039 N/m which is aligned vertically is compressed by 15.416 cm from its relaxed state. A 0.0894 kg mass is then placed on the spring and the spring is released. What is the kinetic energy of the mass as it lifts off from the relaxed spring (see sheet 24)?

V^2=sqrt([(.5*56.039*.15416^2)-(.0894*9.81*.15415)]/(.5*.0894))
V^2=sqrt((.6659-.1352)/.0447
V^2=3.446

KE=(.5*.0894*3.446)
KE=.15402 (-)

I must of did this problem like 4 times!! What am I doing wrong??

Meg
Guest


Back to top Go down

question 2

Post  Meg on Thu Oct 30, 2008 1:23 am

NVM i figured it out..thanks

Meg
Guest


Back to top Go down

Re: Ch 9.3 Help

Post  hello on Thu Oct 30, 2008 11:18 am

number 2 is still giving me a problem..

hello
Guest


Back to top Go down

Re: Ch 9.3 Help

Post  Guest 87 on Thu Oct 30, 2008 3:18 pm

Does anyone know pro. Dawber's email address I looked under staff info and I didnt see anything there.... Thanx

Guest 87
Guest


Back to top Go down

Re: Ch 9.3 Help

Post  1 on Thu Oct 30, 2008 5:01 pm

Question 2
KE= 1/2kx^2-mgx

straight off the lecture notes. the method posted above does not work

1
Guest


Back to top Go down

email address for prof dawber

Post  an on Fri Oct 31, 2008 12:28 am

"Matthew Dawber" <matthew.dawber@stonybrook.edu>

hope that helps

an
Guest


Back to top Go down

Re: Ch 9.3 Help

Post  Guest01 on Fri Oct 31, 2008 2:37 am

The first formula for Question 2 worked for me. Just remember if you find the square root then in the KE formula you have to square the v again. You can also just leave it as is and do .5*m*v (not squared).

Guest01

Posts : 133
Join date : 2008-09-19

View user profile

Back to top Go down

Question 1

Post  Shrubbs on Fri Oct 31, 2008 2:13 pm

Alright, first list all the given quantities

PE = some# (check!)
k = some# (check!)

Now what is the question, what are you looking for? How far has the spring compressed? (In other words, the distance x )

x = ? <------- unknown, which also happens to be the answer


Think of an appropritate equation/formula that would satisfy the problem.

Well, PE = 1/2*k*x^2 pops into mind, don't you agree?

Okay, so now solve for the unknown, the x

So: PE = 1/2*k*x^2
Then: PE / (.5*k) = x^2
Therefore: sqrt [ PE / (.5*k) ] = x

Shrubbs

Posts : 3
Join date : 2008-10-31

View user profile

Back to top Go down

Re: Ch 9.3 Help

Post  Sponsored content


Sponsored content


Back to top Go down

Back to top


 
Permissions in this forum:
You cannot reply to topics in this forum