# Total mechanical energy

1. Nov 5, 2007

### jack1234

I have no clue in solving this question, can somebody help me?
http://tinyurl.com/34ax9f

Last edited: Nov 5, 2007
2. Nov 5, 2007

### malawi_glenn

what have you tried so far?.....

3. Nov 5, 2007

### Staff: Mentor

One has two masses A and B, and the mass of A is twice that of B, i.e. mA = 2*mB.

They are suspended from identical springs and since the mass of A is twice that of B, the deflection of the spring suspending A must be twice that of the spring suspending mass B, because A is twice as heavy.

The force of the spring is F = kx where k is the spring constant and x is the deflection from rest when zero force is applied to the spring.

Now the spring mechanical potential energy is $\int_0^x{F(s)}ds$ = 1/2 kx2, so if the deflection of A is 2x and the deflection of B is x, what can one say about the relationship between the mechanical energies?

Please refer to - http://hyperphysics.phy-astr.gsu.edu/hbase/pespr.html

Last edited: Nov 5, 2007
4. Nov 5, 2007

### jack1234

Hi, I think I am very confused the sign with total mechanical energy....can see here
https://www.physicsforums.com/showthread.php?t=196128

Hence using what I am understand
Treat downward as negative,
for Mass A
E=1/2k(2x)^2 - mg(-2x)
since mg=k(-x), so E=1/2k(2x)^2 - k(-x)(-2x)
E=6kx^2
for Mass B
E=1/2k(x)^2 - mg(-x)
=1/2k(x)^2 - k(-x)(-x)
=3/2kx^2
so E_A=4E_B

Correct...this is the answer...but this will contradict the answer at
https://www.physicsforums.com/showthread.php?t=196128
the answer is -1/2kx^2, not 3/2kx^2(cases for mass B)
What is the problem? I feel very confuse now:(

5. Nov 5, 2007

### jack1234

Hi, learningphysics has very kindly posted a long essay in the mentioned thread, will spend some time to digest it, hope that it will shed me some light for this question :)

6. Nov 5, 2007

### jack1234

Ok, I think the following make more sense after understanding the explanation of learningphysics

for Mass A
(1/2)k(2x)^2-mg(2x)
=2k(x)^2 - k(2x)(2x)
=-2k(x)^2
for Mass B
(1/2)k(x)^2-mg(x)
=1/2k(x)^2-k(x)^2
=-1/2k(x)^2

Hence E_A=4E_B

Is it?

Last edited: Nov 5, 2007
7. Nov 5, 2007

### jack1234

By the way, what I understand from the question is:
Two blocks are hung by two springs separately, ie each block is hung by one spring, not sure is it correct...although the answer is correct.

8. Nov 5, 2007

### learningphysics

yes, looks perfect to me.

9. Nov 5, 2007

### learningphysics

yes, each is balanced by one spring separately...

only difference from what you said is that the masses are on top of springs that are being compressed... but that makes no difference mathematically...

10. Nov 5, 2007

### jack1234

I see, thanks for the confirmation and correction :)

Last edited: Nov 5, 2007
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