How Do I Solve This Work/Energy Problem?

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The discussion revolves around solving a work and energy problem involving gravitational potential energy (GPE) and elastic potential energy (EPE) in a spring. Participants emphasize the conservation of energy principle, stating that the GPE lost by a block sliding down a slope is converted into EPE of the spring. The key equations discussed include GPE (mgh) and EPE (1/2 kx^2), with a focus on determining the vertical distance moved by the block to calculate the change in GPE. Participants also clarify the relationship between forces acting on the block and the spring's tension, leading to the final acceleration equation. The conversation highlights the importance of understanding energy transformations and applying Newton's laws to solve the problem effectively.
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

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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)
 
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I don't get it. I am still confused. Someone please give me a clear explanation. Thanks
 
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?
 
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
 
How much GPE does it lose sliding doen the slope?

If this is converted to EPE of the spring, what is the extension?
 
Well before it falls, the GPE is mgh, where h is xsinθ

but still i dont' know hwo to slove thep roblem
 
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
 
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
muna580 said:
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

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?
 
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  • #11
radou said:
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?

P.S. The displacement down the incline does not equal the displacement in the spring.

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
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
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.

[:smile:]
 
  • #14
Oh okay, how do I do 27?
 
  • #15
I don't understand that. Surely at its lowest point it's come to rest?
 
  • #16
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
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
Yea, i figured that aout, but now I am trying to figure out how to do the next question
 
  • #19
muna580 said:
Yea, i figured that aout, but now I am trying to figure out how to do the next question

Try using Newton's 2nd law.
 
  • #20
radou said:
Try using Newton's 2nd law.


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
 
  • #21
I'll give you the equation in words. :smile:

Set the net force in the direction along the incline equal to ma.
 
  • #22
Wouldn't the net force equal to mgsinθ - T, where T is the tension.

So like you sai,d set the net force equal to ma, is this equation correcy?

mgsinθ - T = ma?
 
  • #23
muna580 said:
Wouldn't the net force equal to mgsinθ - T, where T is the tension.

So like you sai,d set the net force equal to ma, is this equation correcy?

mgsinθ - T = ma?

Yes. You just have to write down what T equals now and plug it into the equation.
 
  • #24
Wait, i think I am a little confused

T = kx

So is it

mgsinθ - kx = ma or is it mgcosθ - kx = ma ?
 
  • #25
muna580 said:
Wait, i think I am a little confused

T = kx

So is it

mgsinθ - kx = ma or is it mgcosθ - kx = ma ?

It is mgsinθ - kx = ma, since we're talking about the direction along the incline. Now divide by m and you'll get acceleration as a function of position x: gsinθ - k/m x = a(x)
 
  • #26
I get a negetive answer. Are you sure its wright?
 
  • #27
Okay, I got the answer right, thanks very much
 

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