Gravitational Potential Energy and Kinetic Energy

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Homework Help Overview

The discussion revolves around a physics problem involving an archer, an arrow, and the concepts of gravitational potential energy and kinetic energy. The original poster presents a scenario where a bow is drawn back, and participants explore the energy transformations involved in launching the arrow.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the types of energy involved, including elastic potential energy and kinetic energy, and question how to relate these to the speed of the arrow. There is confusion regarding the role of the force applied and how it translates into energy.

Discussion Status

Participants have made progress in understanding the relationship between work done by the archer and the kinetic energy of the arrow. Some have calculated the work done and are beginning to connect it to the speed of the arrow, although there remains some uncertainty about the concepts involved.

Contextual Notes

There is a noted lack of clarity regarding the definitions of elastic potential energy versus gravitational potential energy, and how these concepts apply to the problem at hand. Participants express frustration with the complexity of the formulas and the transition between different forms of energy.

tcross420
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Homework Statement


An archer places a 0.15kg arrow on a boasting. Then the archer exerts an average force of 100 Newtons to draw the string back 0.45 meters. Assume that friction is negligible.

A) what speed does the bow give to the arrow? (5 marks)
B) If the arrow is shot vertically upwards into the air, how high will it go? (3 marks)

Homework Equations



I am teaching myself thru adult learning Centre and I have no idea what formulas to use as there are like 10 formulas with many variations.

The Attempt at a Solution



My answer is probably horribly wrong so I'm not even going to post it...unless absolutely necessary.
 
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tcross420 said:

Homework Statement


An archer places a 0.15kg arrow on a boasting. Then the archer exerts an average force of 100 Newtons to draw the string back 0.45 meters. Assume that friction is negligible.

A) what speed does the bow give to the arrow? (5 marks)
B) If the arrow is shot vertically upwards into the air, how high will it go? (3 marks)

So when the force is used to stretch the bow, what kind of energy does it possess? This type of energy is converted to kinetic energy in order to make it move. So can you tell me the equations for these types of energy?
 
Elastic potential energy? It says that this form of energy depends on the mass of the object, acceleration due to gravity and objects height.

If I rearrange the formula for kinetic energy I can get the formula to get the speed. But don't know how to get the kinetic energy to solve the formula. And don't know where the 100 Newtons comes into play.

Ek = kinetic energy
m = mass
V = speed

V = 2 x Ek ÷ m and then square root of that (hopefully you understand)
 
Last edited:
tcross420 said:
Elastic potential energy? It says that this form of energy depends on the mass of the object, acceleration due to gravity and objects height.
That would be gravitational potential energy. Elastic potential energy is another beast, and doesn't involve gravity or heights. Think of energy stored in a spring when it's compressed or stretched away from its equilibrium length.
Still don't really understand how this is going to get me the speed...
When the bow string is pulled back, the archer is doing work and putting energy into the bow. When the string is released, what's going to happen to that energy?
 
The energy would transfer to the arrow. But how do I determine the speed? Does energy = speed?

Gravitational potential energy = Eg

Eg = mass x force x height above reference point
 
tcross420 said:
The energy would transfer to the arrow. But how do I determine the speed? Does energy = speed?
What is the form of energy associated with motion?
Gravitational potential energy = Eg

Eg = mass x force x height above reference point
Not force, gravitational acceleration. The gravitational potential energy, or PEg = m*g*h. But note that m*g is the weight of an object, corresponding to the force due to gravity exerted on the object. So then PEg = weight x height above reference.
 
Energy associated to motion is kinetic energy

Soo..

PEg = m*g*h
PEg = (0.15)(9.81)(0.45)
PEg = 0.66

Sorry i still don't see how this is getting me the speed of the arrow. Thanks for your help and patience I have been out of school for 6 years.
 
tcross420 said:
Soo..

PEg = m*g*h
PEg = (0.15)(9.81)(0.45)
PEg = 0.66

Noooo. There's no gravity involved here. It's elastic potential energy that's involved. The archer is stretching back the bowstring (presumably horizontally).

You were on the right track in your post #3. If you can find the KE then you can find the velocity. The question is, where is the energy coming from that ends up as the KE of the arrow? Hint: a change in KE is the result of work done by a force...
 
I thought you were saying it wasn't elastic potential...and my book says elastic potential depends on mass of object, acceleration due to gravity and objects height. Is my book wrong? It says kinetic energy depends on speed of.object and the mass. But I don't know any speed...

My book says kinetic energy = m*v(squared)÷2 ...I don't have velocity
And the formula to find the speed you need the kinetic energy which I don't know..
 
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  • #10
tcross420 said:
I thought you were saying it wasn't elastic potential...and my book says elastic potential depends on mass of object, acceleration due to gravity and objects height. Is my book wrong? It says kinetic energy depends on speed of.object and the mass. But I don't know any speed...

The bow exhibits elastic potential energy, like a spring. It doesn't involve gravity. Gravity may come into the picture if, for example, a spring has a weight hanging from it. Then gravitational potential energy and spring potential energy will both be involved and energy will move from one to the other. Your book is wrong, or you've taken something out of context, if it states that elastic potential energy alone depends upon gravity. But this is not the case for this problem.

You need to find the source of energy. I gave you a hint about work and kinetic energy. Is there work being done by some force?
 
  • #11
There is 100N of force pulling the string back. I see your hint Hint: a change in KE is the result of work done by a force...

I don't understand that...or don't understand what I'm suppose to do with the 100N
 
  • #12
tcross420 said:
There is 100N of force pulling the string back. I see your hint Hint: a change in KE is the result of work done by a force...

I don't understand that...or don't understand what I'm suppose to do with the 100N

Can you look up the term "work" in your text? It should tell you how to calculate it and describe what it's all about. It should become clear then.
 
  • #13
Work.= force*displacement
Work = 100*0.45
Work = 45 J

How does this help me get the speed? I'm so frustrated. Do I use this answer in the PEg formula?
 
  • #14
tcross420 said:
Work.= force*displacement
Work = 100*0.45
Work = 45 J

How does this help me get the speed? I'm so frustrated. Do I use this answer in the PEg formula?

The above is the work done by the archer on the bow when he pulls back the bowstring. This energy is stored in the bow in the form of elastic potential energy. When the archer let's go of the string, it will put this energy into accelerating the arrow. So the arrow ends up with the energy in the form of kinetic energy. If you go back to your post #3 in this thread you described how you would find the velocity if only you knew the kinetic energy...
 
  • #15
I don't know the kinetic energy tho...unless it's 45 J
 
  • #16
tcross420 said:
I don't know the kinetic energy tho...unless it's 45 J

Yes it is. As I explained in my previous post, the work done by the archer ends up in the bow, then the bow in turn does work on the arrow, moving the energy to the arrow in the from of kinetic energy. So yes, the KE is 45 J.
 
  • #17
Ohhh sorry this is reallllly confusing to me. I got 24.49 m /s for the speed.
 
  • #18
tcross420 said:
Ohhh sorry this is reallllly confusing to me. I got 24.49 m /s for the speed.

That looks fine!

For these sorts of problems, keep an eye out for where the energy is coming from and where it's going to end up. Often all you need to do is find out how much energy is available (in whatever form), and then identify the final form the energy will have when you need to extract your result.

In this case energy went from the archer to the bow, then from the bow to the arrow. You calculated the work done by the archer to find out how much energy went into the bow. Then you made the transition from bow's PE to arrow KE -- a simple trade of energy. From the KE you got your velocity.
 
  • #19
Thanks for all your help, greatly appreciated. So for question b I got 30.58m by using the formula W = m*g*d and rearranged to find d...is this correct?
 
  • #20
tcross420 said:
Thanks for all your help, greatly appreciated. So for question b I got 30.58m by using the formula W = m*g*d and rearranged to find d...is this correct?

Yes that's fine.
 
  • #21
what is the gravitational potential energy of a 60kg student on the surface of earth?
what then minimum energy required to get this student completely out of the Earth gravitational field?


i don't get it, how to solve it.
please help meeeeee
 
  • #22
Mustehsan said:
what is the gravitational potential energy of a 60kg student on the surface of earth?
what then minimum energy required to get this student completely out of the Earth gravitational field?


i don't get it, how to solve it.
please help meeeeee

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