Chemical Potential Energy

In summary, a skateboarder of mass 79 kg, starting from rest at one lip of a half-pipe, moves without friction through one quarter of a circle of radius 6.40 m. After passing point circle b, the skateboarder stands up and raises his arms, lifting his center of mass from 0.500 m to 0.870 m above the concrete at point circle c. Then, he glides upward with his center of mass moving in a quarter circle of radius 6.03 m. When he passes point circle d, his body is horizontal and his speed is 4.66 m/s. To find the amount of chemical potential energy converted to mechanical energy, the conservation of angular momentum is used to
  • #1
flyingpig
2,579
1

Homework Statement



A skateboarder with his board can be modeled as a particle of mass 79 kg, located at his center of mass (which we will study in a later chapter). As shown in the figure below, the skateboarder starts from rest in a crouching position at one lip of a half-pipe (point circle a). The half-pipe is one half of a cylinder of radius 6.90 m with its axis horizontal. On his descent, the skateboarder moves without friction so that his center of mass moves through one quarter of a circle of radius 6.40 m.

p7-59.gif


(b) Immediately after passing point circle b, he stands up and raises his arms, lifting his center of mass from 0.500 m to 0.870 m above the concrete (point circle c). Next, the skateboarder glides upward with his center of mass moving in a quarter circle of radius 6.03 m. His body is horizontal when he passes point circle d, the far lip of the half-pipe. As he passes through point circle d, the speed of the skateboarder is 4.66 m/s. How much chemical potential energy in the body of the skateboarder was converted to mechanical energy in the skateboarder-Earth system when he stood up at point circle b?

Homework Equations



Conservation of Energy

The actual answer is 571J

The Attempt at a Solution



[tex]\Delta[/tex]KE + [tex]\Delta[/tex]PE + [tex]\Delta[/tex]CE = 0

m[v2/2 + g[tex]\Delta[/tex]h] = [tex]\Delta[/tex]CE

79(11.22/2 + 9.8(0.5 - 6.4)] = -[tex]\Delta[/tex]CE

And it isn't 571J...

What in the world is "Chemical Potential Energy"?

Also I don't know what is going on with the tex, but my delta is in lowercase for some reason, so it is supposed to be a triangle

EDIT: I am lagging really bad, so my post may not turn out very pretty
 
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  • #2
hi flyingpig! :smile:

(have a delta: ∆ :wink:)

you've used KE instead of ∆KE :redface:

find ∆KE by finding the new v by using conservation of angular momentum (and why does that work? :wink:)

"Chemical Potential Energy" is is all the energy used up, including that wasted on producing heat and sound

but it doesn't matter … all they're asking is what is the difference in mechanical energy :smile:
 
  • #3
tiny-tim said:
hi flyingpig! :smile:

(have a delta: ∆ :wink:)

you've used KE instead of ∆KE :redface:

find ∆KE by finding the new v by using conservation of angular momentum (and why does that work? :wink:)

"Chemical Potential Energy" is is all the energy used up, including that wasted on producing heat and sound

but it doesn't matter … all they're asking is what is the difference in mechanical energy :smile:

But our class haven't learned angular momentum yet...we are suppose to tackle this problem with only energy and I used ∆KE, energy initially was 0
 
  • #4
hmm … imagine his arms were really long, and he was holding onto the centre of the circle …

when he reaches the bottom, he pulls his arms in, to pull himself up, and his angular speed will increase in exactly the same way as when a spinning skater pulls his arms in (kinetic energy is lost because the skater has to do work to pull his arms in) …

this is conservation of angular momentum …

i don't think you can do the problem without it :confused:

EDIT: actually, you could do it by calculating the work done by the reaction force instead, but that would involve an integration, and you'd need to know the centripetal acceleration formula, v2/r​
 
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  • #5
tiny-tim said:
hmm … imagine his arms were really long, and he was holding onto the centre of the circle …

when he reaches the bottom, he pulls his arms in, to pull himself up, and his angular speed will increase in exactly the same way as when a spinning skater pulls his arms in (kinetic energy is lost because the skater has to do work to pull his arms in) …

this is conservation of angular momentum …

i don't think you can do the problem without it :confused:

EDIT: actually, you could do it by calculating the work done by the reaction force instead, but that would involve an integration, and you'd need to know the centripetal acceleration formula, v2/r​

Okay I know what centripetal acceleration is and that formula. But can you give me some more hints...?I don't know how centripetal acceleration is related to energy or where you getting at
 
  • #6
hi flyingpig! :smile:

centripetal acceleration is related to energy, because it will give you the reaction force, N, and therefore the work done, W …

use F = ma, with a = v2/r and with F = N - mg …

then use W = ∫ N.dh to give you the work done by the reaction force …

ooh, noooo, that's not going to work after all …

you still need conservation of angular momentum to find out what v is (as a function of h)

sorry, looks like you're stuck :redface:
 
  • #7
tiny-tim said:
hi flyingpig! :smile:

centripetal acceleration is related to energy, because it will give you the reaction force, N, and therefore the work done, W …

use F = ma, with a = v2/r and with F = N - mg …

then use W = ∫ N.dh to give you the work done by the reaction force …

ooh, noooo, that's not going to work after all …

you still need conservation of angular momentum to find out what v is (as a function of h)

sorry, looks like you're stuck :redface:

I don't think we will event ouch on angular momentum two weeks after seeing we are still doing linear momentum (1d)

Is it possible to just give me the answer...? Because I've been to khanacademy and got an idea of what angular momentum is, but there is nothing involved in energy. It is just torque
 
  • #8
conservation of angular momentum means that vr is constant, where v is (tangential) speed and r is the radius of curvature (in this case, the distance from the centre of the circle to the centre of mass) …

so when the radius decreases, the speed increases :smile:
 
  • #9
tiny-tim said:
conservation of angular momentum means that vr is constant, where v is (tangential) speed and r is the radius of curvature (in this case, the distance from the centre of the circle to the centre of mass) …

so when the radius decreases, the speed increases :smile:

What does that have to do with energy? We haven't touched on center of mass yet. But why can't mv^2/r dot the arc of the path work?
 
  • #10
flyingpig said:
What does that have to do with energy?

it gives you v, from which you can calculate the energy
We haven't touched on center of mass yet.

but "center of mass" is in the question! :confused:
But why can't mv^2/r dot the arc of the path work?

that would be zero (the two vectors are perpendicular) …

it has to be dot the sudden vertical change in the path :smile:
 
  • #11
Could it be that by raising himself up, he's given himself some potential energy? Had he never raised himself up, he'd only have the potential energy due to his starting position as the initial energy and his energy at the lowest point would be that of just a particle with that specific potential energy. But, by raising himself up, his chemical energy (due to muscles etc) has added an increase in potential energy, thus having more energy at the lowest point than just a particle would have done.
 
  • #12
tiny-tim said:
it gives you v, from which you can calculate the energy

How do I find this constant? It was 0 initially
 
  • #13
hi sungod! :smile:

yes, flyingpig has already included ∆PE (= mgh) in his original equation …

mgh happens to equal the work done (∫ N dh) by N, the vertical reaction force, if the path is flat

but the path is curved, so there's some vertical acceleration, which changes N, which changes the work done to more than mgh :wink:
flyingpig said:
How do I find this constant? It was 0 initially

constant? :confused:

v is the two values of the speed immediately before he stands up and immediately after he stands up
 
  • #14
tiny-tim said:
hi sungod! :smile:

yes, flyingpig has already included ∆PE (= mgh) in his original equation …

mgh happens to equal the work done (∫ N dh) by N, the vertical reaction force, if the path is flat

but the path is curved, so there's some vertical acceleration, which changes N, which changes the work done to more than mgh :wink:


constant? :confused:

v is the two values of the speed immediately before he stands up and immediately after he stands up
0

Oh okay, I think I sort of get it now

V(0) = 11.2m/s
R = 6.4m

V(?) = ?
R = 6.03m

(11.2)(6.4) = x(6.03)

v(?) = 11.88m/s

∆mgh + ∆m/2 (11.88^2 - 11.2^2) = -∆CE

-57.82 + 619.9288 = 562.1088J

Really close now...
 
  • #15
It's due in like two days...please?
 
  • #16
flyingpig said:
V(0) = 11.2m/s
R = 6.4m

V(?) = ?
R = 6.03m

(11.2)(6.4) = x(6.03)

v(?) = 11.88m/s

∆mgh + ∆m/2 (11.88^2 - 11.2^2) = -∆CE

-57.82 + 619.9288 = 562.1088J

(in the exam, you'll need to explain it better than that! :redface:)

Yes, this looks ok :smile:, except I don't see how you got the 57.82.
 
  • #17
tiny-tim said:
(in the exam, you'll need to explain it better than that! :redface:)

Yes, this looks ok :smile:, except I don't see how you got the 57.82.

∆mg(0.5 - 6.4) + ∆m/2 (11.88^2 - 11.2^2) = -∆CE

= 3950J...

I am lost again...please help. it's due tomorrow!
 
  • #18
flyingpig said:
∆mg(0.5 - 6.4) …

oh i see …

no, i think the 0.5 is already subtracted: 6.4 = 6.9 - 0.5

hmm … I've looked at the question again, to see what the second part was, the one with D, and on second reading it looks as if the given speed at D is actually an integral part of finding the chemical energy at the bottom …

it's a weirdly long question, and perhaps you're meant to ignore almost everything in it, and just subtract the KE at A (which is zero anyway) from the KE at D (since the PE at A and D are the same) :confused:
 
  • #19
tiny-tim said:
oh i see …

no, i think the 0.5 is already subtracted: 6.4 = 6.9 - 0.5

hmm … I've looked at the question again, to see what the second part was, the one with D, and on second reading it looks as if the given speed at D is actually an integral part of finding the chemical energy at the bottom …

it's a weirdly long question, and perhaps you're meant to ignore almost everything in it, and just subtract the KE at A (which is zero anyway) from the KE at D (since the PE at A and D are the same) :confused:

I have 4.66^2 * 0.5 * 79 = 857J...
 
  • #20
Quick, back-of-an-envelope kind of thing (I'm taking my zero of energy at the bottom of the original path):
Energy at A = 79*9.81*6.4 = 4954.88
Energy at B = m*(v_b)^2/2 = 4954.88 -> (v_b)^2 = 125.44
Energy at C = m*(v_b)^2 + mg(0.37) = 5246.68
Energy at D = m*(v_d)^2 + mg(6.4) = 5817.69

Energy(D) - Energy(C) = 571

(I'm taking my zero of energy as the bottom of the original path)
 
  • #21
Sorry, I forgot to write down the factor of 1/2 for E(C) and E(D). It's just a typo. The calculation includes the 1/2.
 
  • #22
sungod said:
Sorry, I forgot to write down the factor of 1/2 for E(C) and E(D). It's just a typo. The calculation includes the 1/2.

How did you get 125.4J??

Can you use tex or use these to help me understand?

∫ ≈ ∼ ± ≠ √ ∛ ∜ ∂ ∇ ∆ ≤ ≥ ∼ ∞ ∝ × ÷ ≠ ↑↓↑ ∫ ×10⁻³·×
º ≈ w ≠ ⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ₀ ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ ₉ ⁻³
£ ¢ ¥ € ½ ½ ⅓ ⅔ ¼ ¾ ⅜ ⇌ ➔ ⬅ Å⬆⬇⬌ Ⅽ ⟫ ( ⌘ Å ▼κ ᴺ
ε ω θ ρ Ω µ Ω Δ ∑ π α τ βγδΣΥηλξφϕρσψΔ⌘Å ♢∠⊥
— | ( — ⎍ ⎎ ⌠ ⌡ ·······κΧ∟ ™
≈ ≃ ≈ ≠ ∠ — ▽ Ο
ε ω c ρ Ω µ Ω Δ ∑ π α β γ δ Σ η λ ξ φ ϕ ρ σ ψ ∂ τ
Rʀ Lʟ rᵣ

 
  • #23
sungod said:
Quick, back-of-an-envelope kind of thing (I'm taking my zero of energy at the bottom of the original path):
Energy at A = 79*9.81*6.4 = 4954.88
Energy at B = m*(v_b)^2/2 = 4954.88 -> (v_b)^2 = 125.44
Energy at C = m*(v_b)^2 + mg(0.37) = 5246.68
Energy at D = m*(v_d)^2 + mg(6.4) = 5817.69

Energy(D) - Energy(C) = 571

(I'm taking my zero of energy as the bottom of the original path)

Wait I thuoght at point c, the speed changes
 
  • #24
wait never mind
 

1. What is chemical potential energy?

Chemical potential energy is the energy stored in a substance due to its chemical composition. It is the result of the arrangement and interactions of atoms and molecules within the substance.

2. How is chemical potential energy different from other forms of energy?

Chemical potential energy is a type of potential energy, meaning it is stored energy that has the potential to do work. It is different from other forms of energy such as kinetic energy (the energy of motion) or thermal energy (the energy of heat).

3. What are some examples of chemical potential energy?

Some common examples of chemical potential energy include the energy stored in gasoline, food, batteries, and fireworks. These substances all have the potential to release energy through chemical reactions.

4. How is chemical potential energy measured?

Chemical potential energy is typically measured in joules (J) or calories (cal). Joules are the standard unit of energy in the International System of Units (SI), while calories are commonly used in nutrition and chemistry.

5. How can we use chemical potential energy?

Chemical potential energy can be harnessed to do work in a variety of ways. For example, in a combustion engine, the chemical potential energy in gasoline is converted into kinetic energy to power the vehicle. In a battery, chemical potential energy is released as electrical energy to power electronic devices.

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