# Homework Help: Extension in a spring

1. Feb 26, 2017

### Ameya Darshan

1. The problem statement, all variables and given/known data

Scenario: a block of mass 5 kg is hanging from a spring from an ideal pulley from one side, the other side supporting a mass of 10 kg through a STRING.
Now, the 10 kg block replaced with a 5 kg block.

In which case would the extension in spring be greater, assuming constant acceleration.
2. Relevant equations
10(g-a)=5(g+a) - kx1
and
5(g+a)-kx2= 5(g-a)
3. The attempt at a solution
I know the answer is the first case, visualising it(and according to the solution). But i'm not able to prove this mathematically. From my equations i'm getting x1 proportional to (15a-5g) and x2 proportional to 10a and i don't know how these are related.
Thanks for the help :-)

Last edited: Feb 26, 2017
2. Feb 26, 2017

### kuruman

3. Feb 26, 2017

### Ameya Darshan

my bad i'm so sorry. the spring is connected to the block.

4. Feb 26, 2017

### CWatters

That description doesn't make any sense to me. Any chance of a diagram?

5. Feb 26, 2017

### Ameya Darshan

this is the second case.

6. Feb 26, 2017

### kuruman

Can you find the acceleration in each of the two cases?

7. Feb 26, 2017

### Ameya Darshan

how can i? i mean, it's related to k and extension in the spring. sorry if i'm missing something very common.

8. Feb 26, 2017

### kuruman

Let's look at the second case first with the two 5-kg masses. What do you think the acceleration is in this case?

9. Feb 26, 2017

### Ameya Darshan

i use the equation 5(g+a)-kx = 5(g-a) to get a = kx/10.

10. Feb 26, 2017

### kuruman

Wherever you got that equation, it does not apply here. Suppose I hid the spring with a piece of cardboard (see figure) so that you are not able to see it. All you see is two equal masses on either side of the pulley. You don't know the spring is there and you assume that the string is continuous behind the cardboard. Which way do you think the pulley will be accelerating under these circumstances? Clockwise or anticlockwise?

11. Feb 26, 2017

### Ameya Darshan

wouldn't a be 0 if we assume no spring?

12. Feb 26, 2017

### kuruman

Exactly. It is zero. What is the tension in the left string? What about the right string?

13. Feb 26, 2017

### Ameya Darshan

5g in both.

14. Feb 26, 2017

### kuruman

Very good. Now suppose I remove the cardboard and you discover that there is an extended spring behind it. Would anything change? Would the masses all of a sudden accelerate?

15. Feb 26, 2017

### Ameya Darshan

okay, so the extension will be kx=mg so x =mg/k. am i right?
and was the equation for my first case right?

16. Feb 26, 2017

### kuruman

That is correct.
One thing at a time. I hope you have seen by now that the tension in the string is the force that causes the extension. So now you need to find the tension in the first case.

17. Feb 26, 2017

### haruspex

A massless spring exerts the same force at each end. You seem to have taken it as only exerting a force at one end.

18. Feb 27, 2017

### Ameya Darshan

l don't understand, could you please explain further?

19. Feb 27, 2017

### Ameya Darshan

I get T = 10(g-a) = kx. is that right? but this has 'a' in it's equation, so how would i compare it to the second case?

20. Feb 27, 2017

### haruspex

Ok, but it will be easier if you first explain how you get

21. Feb 27, 2017

### Ameya Darshan

the fact that i'm not able to explain how i got it certainly tells me i was wrong. :-/

22. Feb 27, 2017

### Ameya Darshan

i understood the second case thanks to kuruman but i'm still not able to work out the first case.

23. Feb 27, 2017

### kuruman

Can you draw a force diagram of the left, 10 kg, mass with just a small piece of the string attached to it? Just include part of the string but not the spring. Include, that is, what is showing below the cardboard on the left side of the picture I posted, but with the 10 kg instead of the 5 kg mass. Pretend that you are seeing the mass accelerating down. Can you tell me what forces are acting on it? Remember, you have no idea the spring is there.

Last edited: Feb 27, 2017