betelgeuse1
Posts: 116
Joined: Sat May 09, 2009 10:14 am

I was looking at the practice test and there is problem 2 stating
"Two springs, S1 and S2 have negligible masses and the spring constant of S1 is 1/3 that of S2. When a block is hung from the spring as shown above and the spring comes to equilibrium again the ratio of the work done in stretching S1 to the work done in stretching S2 is
A 1/9
B 1/3
C 1
D 3
E 9"
The answer given by ets is D but I don't really agree with this: I mean, it should be easier (so less work is needed) to stretch the spring with the smaller spring constant so I would say the answer should be 1/3. At least supposing the springs are ideal and obey to the 1/2kX^2 rule for most of the possible dx's

blackcat007
Posts: 378
Joined: Wed Mar 26, 2008 9:14 am

betelgeuse1 wrote:I was looking at the practice test and there is problem 2 stating
"Two springs, S1 and S2 have negligible masses and the spring constant of S1 is 1/3 that of S2. When a block is hung from the spring as shown above and the spring comes to equilibrium again the ratio of the work done in stretching S1 to the work done in stretching S2 is
A 1/9
B 1/3
C 1
D 3
E 9"
The answer given by ets is D but I don't really agree with this: I mean, it should be easier (so less work is needed) to stretch the spring with the smaller spring constant so I would say the answer should be 1/3. At least supposing the springs are ideal and obey to the 1/2kX^2 rule for most of the possible dx's
yes although it will be easier to pull the S1, but the extension in S1 due to the weight of the block (say mg) is 3 times that in S2, the above question can be broken down to the following case. S1 and S2 individually loaded with mg then if x2 if the extension in S2 and x1 in S1 then x1=3*x2 the work done will be 1/2(S1)(x1)^2 and 1/2(S2)(x2)^2 thus in the first case work done is more..
its like pushing a lighter block on a rough floor for a longer distance in comparison to a heavier block for a short distance.. although effort may seem easier in the first case, work done is more.

betelgeuse1
Posts: 116
Joined: Sat May 09, 2009 10:14 am

thanks Christaj
Posts: 2
Joined: Tue Apr 16, 2013 3:37 pm

Blackcat007 is right.
Here's the calculations behind it (I hope I'm correct).
You see the distance extended ($$\vartriangle x$$) by both springs is different. The force however, is the same.
Therefore, $$F_{1} = F_{2}$$

$$- k_{1}\vartriangle x_{1}= - k_{2}\vartriangle x_{2}$$

but $$k_{1}=\frac{1}{3}k_{2}$$

so we have:

$$- (\frac{1}{3}k_{2})\vartriangle x_{1}= - k_{2}\vartriangle x_{2}$$

which simplifies to

$$\frac{1}{3}\vartriangle x_{1} = \vartriangle x_{2}$$

or

$$\vartriangle x_{1} = 3\vartriangle x_{2}$$

now since $$W = \frac{1}{2}k\vartriangle x^{2}$$

$$\frac{W_{1}}{W_{2}}= \frac{\frac{1}{2}k{}_{1}\vartriangle x_{1}^{2}}{\frac{1}{2}k_{2}\vartriangle x_{2}^{2}}=\frac{\frac{1}{2}(\frac{1}{3}k_{2})(3\vartriangle x_{2})^{2}}{\frac{1}{2}k_{2}\vartriangle x_{2}^{2}}= \frac{\frac{1}{3}*9}{1}= 3$$

microacg
Posts: 60
Joined: Sat Mar 03, 2012 2:06 pm

It's nice to help people study for the GRE but you should note that this thread is so old the original poster might have finished their phd by now lol

Christaj
Posts: 2
Joined: Tue Apr 16, 2013 3:37 pm

No need to LOL microacg. Yes, I knew it was an old post, but I was led here because I was taking the exam in April and had come across this old booklet of GRE sample questions and did myself questioned the answer until I came here and understood it better. I improved on the solution a bit better I think. The old booklet GRE file is located here:

Others may be using this resource as well, and may search for an explanatory solution, in which case they may come here and find it. I really did it because of that cause and not the other that made you LOL...

kangaroo
Posts: 130
Joined: Fri Jan 13, 2012 5:31 am