Thread Closed

A safe held by a string breaks and lands on a spring

 
Share Thread Thread Tools
May26-09, 05:39 PM   #1
 

A safe held by a string breaks and lands on a spring


1. The problem statement, all variables and given/known data
A 1209 kg safe is 1.54 m above a heavy-duty spring when the rope holding the safe breaks. Thr safe hits the spring and compress it 53 cm. What is he spring constant of the spring?


2. Relevant equations
K=mV^2/2
U=mgh
Ki+Ui=Kf+Uf


3. The attempt at a solution

I thought I understood momentuem and energy but not anymore. Here is what I did....

Ui=mgh= 2*9.8*1.54
Kf=mV^2/2

Then solved for V and got 5.494 m/s

Then used that velocity for the spring constant by using......

U=k(delta x)^2/2
K=mV^2/2

then solved for k and got 129912 N/m...... but that is coming up wrong what am i doing wrong.
 
PhysOrg.com
PhysOrg
science news on PhysOrg.com

>> Hong Kong launches first electric taxis
>> Morocco to harness the wind in energy hunt
>> Galaxy's Ring of Fire
May26-09, 06:28 PM   #2
 
any ideas?
 
May26-09, 06:33 PM   #3
 
Recognitions:
Gold Membership Gold Member
Homework Helper Homework Help
Science Advisor Science Advisor
Well for one thing you are using m = 2 kg when it is given that m = 1209 kg. Then you neglect the gravitational potential energy in the 2nd set of equations. It is better to do this all in one step: Initial energy = Final energy, where v_initial = v_final = 0.
 
May26-09, 06:34 PM   #4
 
Blog Entries: 27
Recognitions:
Gold Membership Gold Member
Homework Helper Homework Help
Science Advisor Science Advisor

A safe held by a string breaks and lands on a spring


Hi talaroue!

(try using the X2 and X2 tags just above the Reply box )
Quote by talaroue View Post
A 1209 kg safe is 1.54 m above a heavy-duty spring when the rope holding the safe breaks. Thr safe hits the spring and compress it 53 cm.

Ui=mgh= 2*9.8*1.54
No, the safe stops moving 53cm lower.
 
May26-09, 06:36 PM   #5
 
JAY:

so your saying that i should just use U=mgh and U=k(delta x)^2/2....solve for k and get....

k=2mgh/delta x^2

TINY TIM:

so I didn't go about the problem wrong I just have to add the .53 m to the hieght?
 
May26-09, 06:42 PM   #6
 
Admin
The safe starts 1.54 m (154 cm) above the spring and then travels another 0.53 m after contact. That must be considered with respect to gravitational potential energy.
 
May26-09, 06:46 PM   #7
 
Recognitions:
Gold Membership Gold Member
Homework Helper Homework Help
Science Advisor Science Advisor
Quote by talaroue View Post
JAMES:

so your saying that i should just use U=mgh and U=k(delta x)^2/2....solve for k and get....

k=2mgh/delta x^2
Yes, but 'h' is as noted by others, and don't forget to correct your value of 'm'.
 
May26-09, 06:47 PM   #8
 
ahhh I see at least I had the right idea just need to be more careful. If you guys don't mind I am having problems with another problem as well.... it might be a same problem i am going to go back and look at it but I made a thread about it as well....

Second problem I have issues with
 
May26-09, 06:50 PM   #9
 
Quote by PhanthomJay View Post
Yes, but 'h' is as noted by others, and don't forget to correct your value of 'm'.
I wasn't using 2 as my m, that was just noting the second equation is U=k(delta x^2)/2 I didn't combined them when i posted that, i see what you are saying though.
 
Thread Closed
Thread Tools


Similar Threads for: A safe held by a string breaks and lands on a spring
Thread Forum Replies
Block held against a spring going up a frictionless plane Introductory Physics Homework 3
Tension in a String just before it breaks (Circular Motion) Introductory Physics Homework 4
[String-schools] Spring program in Italy Beyond the Standard Model 0
[String-schools] Spring program in Italy Beyond the Standard Model 0
String Theory: Forever safe? Beyond the Standard Model 13