# Homework Help: Work/Engery Problem: Emergency

1. Nov 1, 2006

### muna580

I am trying to do this problem but I have no idea howa to do it. Can someone please give me a step by step explanation of how to do it? Please, thanks

http://img351.imageshack.us/img351/3290/untitlediq5.png [Broken]

Last edited by a moderator: May 2, 2017
2. Nov 1, 2006

### rsk

hang on, rethinking

On second thoughts I'd say use energy.

How much GPE does it lose sliding doen the slope?

If this is converted to EPE of the spring, what is the extension?

(First method, using force, gives me an answer eactly half of this .... trying to work out why)

Last edited: Nov 1, 2006
3. Nov 1, 2006

### muna580

I don't get it. I am still confused. Someone please give me a clear explanation. Thanks

4. Nov 1, 2006

Or, you can use an energy approach, since I assume that was the topic of your last unit. Did you learn something about conservation of energy?

5. Nov 1, 2006

### muna580

Yes, this homework relateds to the Work, Engery stuff. I know that the potetnial engery is mgh, and kenetic engery is 1/2mv^2. Also, for springs, teh engery is 1/2kx^2. But I don't undertsand how to eo this problem

6. Nov 1, 2006

### rsk

How much GPE does it lose sliding doen the slope?

If this is converted to EPE of the spring, what is the extension?

7. Nov 1, 2006

### muna580

Well before it falls, the GPE is mgh, where h is xsinθ

but still i dont' know hwo to slove thep roblem

8. Nov 1, 2006

### rsk

No - just look at the change in GPE.

What distance (vertical) does it move? This allows you to find the change in GPE, which must be converted to EPE

9. Nov 1, 2006

### muna580

If you make the ground level to be the stoping point of the block, then we can say that

sinθ = h / x

....wait, nvm, i am confused, can you please give me a detalted explanation

10. Nov 1, 2006

Define your system - it consists of a spring and a block. The energy of the system is defined as the sum of kinetic and potential energy, for each part of the system. Energy is conserved. Take two points, 1 and 2. Let point 1 be the point when the block is held, and it posesses gravitational potential energy only. Let point 2 be the point when the block has sled down, and stopped. At that point it posesses no kind of energy, but the spring posesses elastic potential energy.

Since energy is conserved, the energy at point 1 must equal the energy at point 2. What equation does that give you?

Last edited: Nov 1, 2006
11. Nov 1, 2006

### muna580

So, at point 1, where the block is held initially, I got the following engery equation

Point 1

(1/2)mv^2 + mgh = (1/2)kx^2
0 + mgh = 0
mgh = 0

0 for kenetic energy because the block is not moving, and 0 for spring because it is not yet stretched.

Point 2

(1/2)mv^2 + mgh = (1/2)kx^2
0 + 0 = (1/2)kx^2

0 for potential enegery because it has reached OUR made up groud level
0 for kenetic enegry because we it stops at this pont.

So, what do I do now?

12. Nov 1, 2006

### rsk

Why is mgh at point 1 zero? It's above your 'ground' level

What vertical distance h does the block move? (use x and trigonometry)

Now, use that number to find the change in GPE=mgh

Now put that = EPE in the spring.

13. Nov 1, 2006

Please ignore what I said about the displacement down the incline not being equal to the displacement in the spring. It was pretty stupid, and I deleted it.

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14. Nov 1, 2006

### muna580

Oh okay, how do I do 27?

15. Nov 1, 2006

### rsk

I don't understand that. Surely at its lowest point it's come to rest?

16. Nov 1, 2006

### muna580

Thats what I am thinking if they are saying the lowest point. But I don'twant to risk putting in the answer as 0 because i am pretty sure they are asking something else.

What is they are asking you for the acceleartion right before it stop? How do I find the acceleartion fight before it stops?

17. Nov 1, 2006

Ok, once again:

mg x sin(30) = 1/2 k x^2.

The term on the left side is the energy of the system at point 1, and the term on the right side is the energy of the system at point 2. I hope this is clear now.

18. Nov 1, 2006

### muna580

Yea, i figured that aout, but now I am trying to figure out how to do the next question

19. Nov 1, 2006

Try using Newton's 2nd law.

20. Nov 1, 2006

### muna580

can you please give me the equation? Please? I only have 20 mins to submit thsi answer. please, thanks after the homework, i will pick up the solutions and see how they got the answer. please