# Desperately need help on potential energy problem

#### rasgar

need help on potential energy problem (less than 45 min left))

This is due online at 11:00 pm tonight! Otherwise I wouldn't post here I'd ask my prof for help.

1. The problem statement, all variables and given/known data
A 20.0 kg block is connected to a 30.0 kg block by a string that passes over a light, frictionless pulley. The 30.0 kg block is connected to a spring that has negligible mass and a force constant of 280 N/m, as shown in the figure below. The spring is unstretched when the system is as shown in the figure, and the incline is frictionless. The 20.0 kg block is pulled 20.0 cm down the incline (so that the 30.0 kg block is 40.0 cm above the floor) and released from rest. Find the speed of each block when the 30.0 kg block is 20.0 cm above the floor (that is, when the spring is unstretched). P.S. the incline is 40 degrees above the horizontal and the 20kg weight is on that horizontal, while the 30kg block is vertical.
2. Relevant equations
.5mV^2=PE
.5Kx^2=PE
KE=PE
3. The attempt at a solution
I set up the problem to have potential energy of the 30kg block increased and the 20kg block decreased, with the potential energy in the spring increased. This came out to 30*9.81*.2-20*9.81*.2/sin(40)+.5*280*.2^2. With the total potential energy of the system I promptly set it equal to the kinetic energy and solved for the mass of the system (sqrt(2*PE/50)). The answer was not correct. I looked in my text book for a similar question and found one exactly like it, except with the spring constant being 250. The back of the book said the answer was 1.24m/s. I tried to replicate the results with my previous attempt with no avail. And I've tried all deviations, such as changing the plus and minuses, and solving for individual masses. Got close but no cigar. The book and online homework said specifically to carry out all operations to 4 decimal places, and I did, so that's not the problem either. For the love of god help me quickly.

Last edited:

#### Oerg

I have no idea what you are talking about maybe you should attach a picture

#### rasgar

heres a digram attached. Also, I've tried setting the change in kinetic energy to incorporate the fact that the 20kg wight goes along the incline, so the new attempt has sqrt(2*PE/(30+20/sin(40))). After playing around with it for a while, still no cigar.

#### Attachments

• 9.1 KB Views: 291
Last edited:

#### Oerg

i really have no idea what is attached to what the question is so vague without a diagram

#### rasgar

Sorry it says the attachment is pending approval. Here's a direct link.

http://i72.photobucket.com/albums/i185/panda****er/p7-45.gif [Broken]

Also, the profanity editor is preventing the link from being posted. I know the name is childish, but this is my photobucket account from when I was around 13 and I just revived it today. So where you see the 4 *, just put the f-word in there (with a capitol F, since it's case sensitive)

Last edited by a moderator:

#### Tedjn

Link isn't working. Is the URL being filtered?

#### Oerg

OHHHH so the spring is connected to the ground....

#### Oerg

you should calculate the position of the spring when no weight are attached. I found that to be about 0.6m. So in the diagram, the spring's potential energy at that moment is calculated according to the extension which is 0.4m. This is where i thought you went wrong.

So now you can calculate the potential energy before it was released and then do some arithemic to find the kinetic energy at the equilibrium position.

EDIT: oh yeh why do you want to f*** a panda when you were* 13

Last edited:

#### rasgar

I don't think that helps. The problem said that the spring is unstretched, and I'm pretty sure it's not compressed. My results for the calculation were too low, so removing even more potential energy from the system would make the situation worse. Please note that if you think you have it, please use the 250N/m spring constant and see if you get 1.24m/s.

edit: i thought the name was funny at the time. I've obviously grown out of it.

Last edited:

#### rasgar

I have 1 hour and 15 minutes as of the end of this post. Please hurry. I'm not lazy just in a rush.

#### Oerg

hi im sorry the extension should be 0.6m.

So invoking the law of conservation /energy gives us

P.E. Elastic + P.E. Grav When height of Spring =0.2m

-

P.E. Elastic + P.E. grav when height of spring=0.4m

=

K.E.

of course this is assuming that the psring was compressed

#### rasgar

There's about 15 minutes left and that still doesn't fix it. Try to solve it and see if you get a speed of 1.24m/s. Taking away potential energy will make the system go even slower than my already too slow answer.

#### rasgar

Update: Since the homework system allowed me 3 guesses before I get marked off, I took a stab at it. Since the spring constant is only 30N/m, and the problem calls for a small change in distance, I figured it wasn't too far away form 1.24. The answer is 1.26, but I still don't know how to get there. Can someone please tell me?

edit: never mind I'll ask my professor. Thanks for you effort anyways guys.

Last edited:

#### Oerg

I managed to solve the question through use of newtonian laws of motion.

First calculate the average force the spring would exert on the masses + the weight the two boxes are exerting that would cause them to accelerate.

Then you can use simple equations to solve for the final velocity.

#### Oerg

but the above was using spring consant of 250, if i used 280 i got 1.25

#### Oerg

i also got the result, 1.25 using conservation of energy in a closed system with spring constant 250.

### 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