Gravitational Potential Energy homework

[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...

 

[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.

 

[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".

 

[Doc Al]

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".

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.

 

[freshbox]

I thought there is height at datum 1?

 

[freshbox]

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

 

[Doc Al]

I thought there is height at datum 1?

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.

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

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.

 

[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.

 

[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!

 

[Doc Al]

Doc Al: If i choose my reference point at datum 1, = 0. My GPE at datum 2 = -ve since it's lower.

Right. No problem.

Then how do i calculate my GPE = mgh? It is a horizontal line, there isn't any height at all.

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.)

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.

If energy is conserved and the GPE decreases then the other energy forms must increase.

 

[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?

 

[Doc Al]

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 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.)

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?

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.

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?

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.)

 

[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.Please advice, thanks.

 

[Doc Al]

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

 

[freshbox]

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

 

[Doc Al]

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

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

Fill in those blanks.

 

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

 

[Doc Al]

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

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

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?

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

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.

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

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.)

 

[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

 

[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.

 

[Doc Al]

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

One thing to get straight is what value of v are you using? You want the v at datum 2. (Of course, you can just as easily use the energy at datum 1 and save yourself a calculation.)

And it would be a good idea to rewrite your values from post #17 using clearer notation. Be sure to include v and the work done by friction.

 

[freshbox]

I am using v = 4.36m/s obtained from datum 2.

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

"Work done by friction" is from datum 2 to 3?

 

[Doc Al]

I am using v = 4.36m/s obtained from datum 2.

I thought that was the answer for question a, which is at the point marked X not at datum 2. (It's 2 m from datum 2.)

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

"Work done by friction" is from datum 2 to 3?

If you are comparing datum 2 to datum 3, then yes.

 

[freshbox]

let me think abit

 

[freshbox]

datum 2
-----------
E2=K2+G2+S2
E2=1/2(8)v^2+8x9.81x-sin25x3+0
=4v^2-99.50

So after calculations I found out v = 4.36m/s

Since E2=4v^2-99.50, sub v = 4.36 in
E2=-23.4616

This is what i meant. Is there anything wrong with my datum 2 equation? I am going to compare with datum 3.

 

[Doc Al]

This is what i meant. Is there anything wrong with my datum 2 equation? I am going to compare with datum 3.

There are at least 5 points of interest in this problem:
- datum 1: top of the ramp
- datum 2: bottom of the ramp
- the point marked X
- the initial position of the uncompressed spring
- the point of maximum compression

The speed you calculated was for the third point on this list, which is not datum 2. But you can certainly use it.

 

[freshbox]

Thank you Doc Al for the help.