# Homework Help: Elastic potential energy problem help

1. Nov 5, 2007

### jack1234

[SOLVED] Elastic potential energy

I am not sure what the question ask, and have no clue in solving this question, can somebody help me?
http://tinyurl.com/3a8vnu

2. Nov 5, 2007

### mjsd

so you basically have one long spring with total stretch of 2x, don't you think?

3. Nov 5, 2007

### jack1234

yes, I have thought of this,
so...it should be mg2x=2mgx
but the answer is mgx...hence now I am totally lose.

4. Nov 5, 2007

### mjsd

note, the question ask you for potential energy... so it IS mgx

5. Nov 5, 2007

### jack1234

Thanks, but Is it possible to explain how to get mgx? What I am thinking is the spring has extended 2X, hence the change in height of the object should be 2X...

6. Nov 5, 2007

### learningphysics

The way I did it is: you have 2 springs with constant k each stretches by x.

elastic pot. energy = 2*(1/2)kx^2 = kx^2

we also know that by the hanging mass freebody diagram. mg = kx. so k = mg/x

plug k into the above elastic pot. energy equation...

7. Nov 5, 2007

### mjsd

alternatively, replacing x above with 2x everywhere and take away the "2*", and treat it as one big string also gives you the same result.

8. Nov 5, 2007

### jack1234

Very thanks!

Am I right that the following
decrease in gravitational potential energy(=2mgx)=increase in elastic potential energy
does not work because it can be
decrease in gravitational potential energy=increase in elastic potential energy + kinetic energy?

9. Nov 5, 2007

### learningphysics

yes... don't use a gravitational pot. energy approach here because that leads to an oscillating mass... but that's not what we're dealing with here... we're dealing with a mass hanging at rest...

mjsd's approach is fine. I just want to note that the spring constant of the combined spring is different thatn the spring constant of each of the two springs...

the hanging mass:

kx = mg (where k is the spring constant of 1 spring)

k = mg/x

kcombined*(2x) = mg (where kcombined is the spring constant of the 2 joined springs)

kcombined = mg/2x = k/2.

so using the combined spring:

mg = kcombined(2x)

kcombined = mg/(2x)

el. pot energy = (1/2)kcombined*(2x)^2 = (1/2)(mg/(2x))*(2x)^2 = mgx

10. Nov 5, 2007

### jack1234

Thanks :)...but unfortunately I got confused with the sign because the following question
http://tinyurl.com/3a9gjg
the answer for this question is b (negative value)

May I know why the the question above has negative value of mechanical energy, while this question has positive value of elastic potential energy?
How do we decide for the sign?

11. Nov 5, 2007

### learningphysics

the elastic potential energy was positive because we treat the 0 stretch position as having 0 energy. and we were only dealing with elastic potential energy in that problem (the question asked for elastic potential energy, not mechanical energy or gravitational potential energy).

here we have elastic potential energy + gravitational potential energy... again we set the 0 stretch position as being 0 elastic potential energy... and we set a particular height as 0 gravitational potential energy. below that height, gravitational potential energy is negative.

gravitational potential energy = mgh, where h is the displacement from the 0 energy position...

if the spring stretches by x, that means the mass hangs x distance below the 0 energy position... what is the gravitational potential energy. what is the elastic potential energy?

12. Nov 5, 2007

### jack1234

Thanks, for
http://tinyurl.com/3a9gjg
if we take downward as positive
elastic potential energy + gravitational potential energy
=1/2kx^2 + mgx
=1/2kx^2 + kx^2 (mg=kx)
=3/2kx^2

if we take downward as negative
elastic potential energy + gravitational potential energy
=1/2kx^2 + mg(-x)
=1/2kx^2 + k(-x)(-x)
=3/2kx^2

non of the result leads to -1/2k^2...what is the problem?

13. Nov 5, 2007

### learningphysics

if the object hangs x distance below the 0 position. then gravitational potential energy = -mgx not mgx.

14. Nov 5, 2007

### learningphysics

mg = kx

so:

(1/2)kx^2 + mg(-x)
=(1/2)kx^2 + kx(-x)
=(1/2)kx^2 - kx^2
= -(1/2)kx^2

15. Nov 5, 2007

### jack1234

Thanks a lot, I think this is the crux I don't understand, and I am very eager to get it clear :)

If we take downward as positive why it can't be mg(+x)?

If we take downward as negative, why it is not k(-x)(-x)?

16. Nov 5, 2007

### learningphysics

We can choose the directions for position... and we can choose the 0 energy position... but after that, gravitational potential energy is fully determined... we can't choose beyond this point...

we can choose the positive direction and negative direction for position... but after that we can't choose when it comes to energy... we can choose a reference energy and that's it.

suppose we take downwards as positive. x>0.

and suppose we choose energy at x = 0, to be 0.

now gravitational potential energy at all other points is full determined.

so what is the gravitational potential energy at x=x1, where x1>0

by definition of potential energy... it is the negative of the work done by gravity in moving from x = 0 to x = x1.

gpe at x =x1 is:

$$-\int_0^{x_1}{\vec{F}\cdot\vec{dx}$$
=$$-\int_0^{x_1}{mg*dx}$$ (notice that mg acts downwards so it is positive.)
= -mgx1

suppose instead we took downwards as negative and upwards as positive... we want the gravitational potential energy at the same position... but then the same position is located at x = -x1 .

now we need gpe at x = -x1

$$-\int_0^{-x_1}{\vec{F}\cdot\vec{dx}$$
=$$-\int_0^{-x_1}{-mg*dx}$$ (now mg is negative since downwards is negative here.)

= $$\int_0^{-x_1}{mg*dx}$$
= -mgx1

so what's the main point here... no matter what coordinate system we use... the gravitational potential energy at a point below the 0 energy position will be negative. Same way no matter what coordinate system we use, the gravitational potential energy at a point above the 0 energy position will be positive.

suppose we have some 0 position x = 0... now suppose we take up positive down negative... we have some position x =x1.

what is the gravitational potential energy at x=x1...

it is:

0 + negative of work done by gravity to go from 0 to x1

0 + [-(-mg)(x1-0)] = mgx1

if we switch to up negative downpositive. now we want potential energy at a point x = x1 (using the new coordinate system).

0 + negative of work done by gravity to go from 0 to x1

0 + [-mg(x1-0)] = -mgx1

but remember... these give the same results... because a position of x =x1 in the old coordinates has coordinates of x = -x1 in the new coordinates... leading to the same gravitational potential energy!

In everything below, I'm assuming that the mass hangs below the 0 position...

if we take down as negative... and suppose x>0... that means that the position is actually -x in this coordinate system... spring force = -k*displacement (this is always true)

spring force + gravitational force = 0

-k(-x-0) - mg = 0
kx = mg.

what is the total energy... -mgx + (1/2)kx^2 = -(1/2)kx^2

suppose we take up as negative... down as positive... and again x is given as >0... this time the position is x in this coordinate system.

-k(x-0) + mg = 0
kx = mg

what is the total energy... -mgx + (1/2)kx^2 = -(1/2)kx^2

Now... suppose x<0 as it is given... (ie they're giving a negative number). suppose we take down as negative up as positive... we can leave x as it is because it fits with the coordinate system...

-k(x-0) - mg = 0
-kx = mg

what is the total energy at this x value... mgx + (1/2)kx^2 = -kx^2 + (1/2)kx^2 = -(1/2)kx^2

suppose x<0 as it is given. suppose we take positive down negative up... we need to switch to -x because it is below 0 (we assume this from the context of the problem)...

-k(-x-0) + mg = 0
-kx = mg

what is the total energy at this x value... mgx + (1/2)kx^2 = -kx^2 + (1/2)kx^2 = -(1/2)kx^2

lol! Looks like I rambled on a lot... hope what I wrote makes sense.

A main point... the x could be positive or negative... but as long as the actual position is below the 0 energy position... we get a result of -(1/2)kx^2.

17. Nov 5, 2007