Desperately on potential energy problem

In summary: So I solved for the potential energy before the release and then used the Pythagorean theorem to find the kinetic energy. I got 1.24m/s. Sorry it's not much better.In summary, I am having trouble with a potential energy problem and I need help.
  • #1
rasgar
32
0
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.

Homework Statement


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.

Homework Equations


.5mV^2=PE
.5Kx^2=PE
KE=PE

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 textbook 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.
 
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  • #2
I have no idea what you are talking about maybe you should attach a picture
 
  • #3
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

  • p7-45.gif
    p7-45.gif
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  • #4
i really have no idea what is attached to what the question is so vague without a diagram
 
  • #5
Sorry it says the attachment is pending approval. Here's a direct link.

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

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)
 
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  • #6
Link isn't working. Is the URL being filtered?
 
  • #7
Read the edit
 
  • #8
OHHHH so the spring is connected to the ground...
 
  • #9
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
 
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  • #10
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.
 
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  • #11
I have 1 hour and 15 minutes as of the end of this post. Please hurry. I'm not lazy just in a rush.
 
  • #12
hi I am 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
 
  • #13
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.
 
  • #14
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.
 
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  • #15
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.
 
  • #16
but the above was using spring consant of 250, if i used 280 i got 1.25
 
  • #17
i also got the result, 1.25 using conservation of energy in a closed system with spring constant 250.
 

Related to Desperately on potential energy problem

1. What is potential energy and why is it important?

Potential energy is the energy an object possesses due to its position or configuration. It is important because it allows us to understand how energy is stored and transferred in different systems, and can be harnessed for practical use.

2. How does potential energy relate to the concept of work?

Potential energy and work are closely related, as work is the transfer of energy from one form to another. In the case of potential energy, work is done when an object's position or configuration changes, resulting in a change in its potential energy.

3. Can potential energy be converted into other forms of energy?

Yes, potential energy can be converted into other forms of energy such as kinetic energy, thermal energy, or electrical energy. This conversion can occur naturally or through human intervention.

4. What factors affect the amount of potential energy an object possesses?

The amount of potential energy an object possesses depends on its mass, height, and the gravitational force acting upon it. The higher the object is located and the greater its mass, the more potential energy it will have.

5. How can potential energy be calculated?

The potential energy of an object can be calculated with the formula PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object. Alternatively, potential energy can also be calculated by multiplying the force applied to an object by the distance it moves in the direction of the force.

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