# Homework Help: Gravitational Potential Energy homework

1. Jul 8, 2012

### freshbox

datum 1
---------
E1=K1+G1+S1
=0+8(9.81)(-3)(sin37)+0
=-141.69

datum 2
---------
E2=K2+G2+S2
=1/2(8)(v^2)+0+0
= 4v^2

Frictional Force: 0.15x9.81.8 = 11.772Newtons

U1-2= -11.722x2
= -23.544J

E1+U1-2=E2
-144.69-23.544=4v^2
-42.0585=v^2
Can't square root a -ve number

Something is wrong somewhere....

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2. Jul 8, 2012

### Undoubtedly0

First, why is G1 negative? Using your choice of zero gravitational potential energy, is not G1 = mgh, where h = 3sin(25°)? A mass above the line of zero gravitational potential energy will have positive gravitational potential energy.

3. Jul 8, 2012

### freshbox

Because the values I took are behind datum 1. From my textbook "all distances above the datum are taken as positive while distances below the datum are taken as negative".

4. Jul 8, 2012

### Staff: Mentor

Are you measure GPE from the position of 'datum' 1? If so, its GPE would be 0 and the GPE of datum 2 would be negative.

You must be consistent in what you use for a reference point.

5. Jul 8, 2012

### freshbox

I thought there is height at datum 1?

6. Jul 8, 2012

### freshbox

I thought when a object is hanging or in midair, there is GPE.

7. Jul 9, 2012

### Staff: Mentor

Sure. But measured with respect to what reference point? Using its own position as y=0, its height would be zero. Up to you which height you choose as your reference point. Pick one and stick to it.
Sure, but that depends on where you choose to set GPE = 0.

For this particular problem, I would choose the height of the horizontal portion as my GPE = 0 reference. Thus the GPE at datum 1 would be positive and the GPE at datum 2 would be zero.

8. Jul 9, 2012

### HallsofIvy

Potential energy is always relative to some arbitrary reference point. Here, you can choose that reference point to be datum 1 or datum 2. If you choose datum1 as reference point, the potential energy there will be 0 and the potential energy at datum2, which is lower, will be negative. If you choose datum2 as a reference point, the potential energy there will be 0 and the potential energy at datum1, which is higher, will be positive. In either case, the change in potential energy will be negative so, in order to keep total energy constant, the total of the change in kinetic energy and energy lost to friction must be positive.

9. Jul 9, 2012

### freshbox

Doc Al: I got the answer by setting datum 1 as reference point. Let's say if i set my reference point at datum 2, will I get the same answer?

HallsofIvy: "the change in potential energy will be negative so, in order to keep total energy constant, the total of the change in kinetic energy and energy lost to friction must be positive." I don't quite understand this sentence, can you please rephrase it.

Thanks alot!

Last edited: Jul 9, 2012
10. Jul 9, 2012

### Staff: Mentor

Right. No problem.
When using GPE = mgh, the h is the 'height' with respect to your reference point. If your reference is at datum 1, then the height at datum 2 is negative. (You might find it easier to think of mgh as mgΔh or mgΔy.)
If energy is conserved and the GPE decreases then the other energy forms must increase.

11. Jul 9, 2012

### freshbox

Ok i just verified i got the same answer setting the datum the other way round. Thank you so much for the help. Out of curiosity, normally do we set the datum as -ve or +ve?

I set datum 1 as reference point, so GPE at datum 2 is -ve. You said "If energy is conserved and the GPE decreases then the other energy forms must increase" and since my GPE at datum 2 is -ve (decrease) how come my frictional force in my calculation is -ve. Hmm?

For Part B, I set my datum at position 3. So there are 3 datum point now. Should I still take reference from datum 1?

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12. Jul 9, 2012

### Staff: Mentor

I don't know what you mean by 'set the datum as -ve or +ve'. It's up to you to choose a point to call GPE = 0. Once you choose that point, the GPE at other points is set. (And that could be negative or positive, depending on whether other points are higher or lower than your reference level.)

It's somewhat a matter of semantics. The energy 'lost' to friction is positive, but the work done by friction is negative. Here's how I would express it:
ΔKE + ΔGPE + ΔSPE = Work done by friction

Since the work done by friction is negative, the mechanical energy decreases as the block slides over the rough segment.

The important thing is that when you are comparing two datum points to set up an energy equation, you must use the same reference point for GPE.

You can use datum 1 as your reference.

My personal preference is to always choose the lowest point as my GPE = 0 reference point. Then all values of GPE are 0 or positive. So I would have chosen the height of datum 2 as my GPE = 0 point. But it doesn't matter. (You'll get the same answer either way.)

13. Jul 9, 2012

### freshbox

datum 3
----------
E3=K3+G3+S3
=0-99.50+1/2(5000)(x)^2
=2500x^2-99.50

F=11.772 Newtons (As Caculated from my previous working)
E2=4v^2-99.50 (As Caculated from my previous working)
V=4.36m/s
Hence E2=-23.4616

U2-3=FxS
=-11.772x(5+X)
=-58.86-11.772X

E2+U2-3=E3
-23.4616-58.86-11.772X=2500x^2-99.50
=2500x^2+11.772x-17.1784

Using Quadratic Formula to solve: Ans: 0.0825 which is wrong.

Did I go somewhere wrong in the U2-3 part? I am thinking whether i should use U1-3 or U2-3.

Last edited: Jul 9, 2012
14. Jul 9, 2012

### Staff: Mentor

It's a bit unclear to me what two points you are comparing. Please define them clearly.

15. Jul 9, 2012

### freshbox

I am comparing datum 2 and datum 3. Taking GPE reference point from datum 1 which is -99.50 obtained from datum 2.

16. Jul 9, 2012

### Staff: Mentor

OK, so your energy equation should be:
(KE3 + GPE3 + SPE3) - (KE2 + GPE2 + SPE2) = Work done by friction

Fill in those blanks.

17. Jul 9, 2012

### freshbox

Is (KE3 + GPE3 + SPE3) - (KE2 + GPE2 + SPE2) = Work done by friction

derive from E2+U2-3=E3
E3-E2=U2-3 ?

If yes

(KE3 + GPE3 + SPE3) - (KE2 + GPE2 + SPE2) = Work done by friction
(0 + 8x9.81xsin25x-3 + 1/2(5000)(x)^2) - (1/2(8)v^2 + 8x9.81xsin25x-3 + 0) = Work done by friction

And if my blanks are correct, is Work done by friction = U2-3?

Last edited: Jul 9, 2012
18. Jul 9, 2012

### Staff: Mentor

It's a restatement of the energy equation I gave in post #12.

I don't quite understand your notation. What's E and U? Is E kinetic energy and U potential? I don't know what the -3 refers to. Do you mean U2 - U3?

Careful with your notation! You are using 'x' to represent both multiplication and the amount by which the spring is compressed. Not a good idea. (Note that the two GPE terms cancel out, since your two points are at the same height and thus have the same GPE.) Use 'x' to represent the spring compression only; use * (or nothing) to represent multiplication. And use parentheses as needed to avoid confusion.
The work done by friction is the friction force times the displacement. Express that displacement in terms of x. (Keep in mind that the work done by friction will be negative.)

19. Jul 9, 2012

### freshbox

WORK OF A FORCE
When a force acts on a body and causes it to move a distance from position 1 to position 2, then work is done on the body.

The work of a force can be written in mathematical form as

U1-2=Fs

where U1-2 is the work done
F=force
s=distance

WORK ENERGY PRINCIPLE
A body or a system of bodies at any instant has a sum of energies of kinetic energy, gravitational potential energy and elastic potential energy.

The sum of energies is given by E=K+G+S

When a bodies moves from initial situation 1 to another situation 2, the work energy principle states that the energy at situation 2 is equal to the energy at situation 1 plus the work of external forces acting from situation 1 to 2.

The work energy PRINCIPLE can be written as E1+U1-2=E2

20. Jul 9, 2012

### freshbox

the x I got is 42.605. Even bigger. can you help me take a look whether the values I put are correct please.