Energy Conservation of a toy gun

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Homework Help Overview

The discussion revolves around the energy conservation principles applied to a toy gun that fires a projectile disc. The problem involves calculating the speed of the disc as it exits the barrel, taking into account the spring's potential energy, frictional forces, and kinetic energy.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of energy conservation equations, including kinetic energy, potential energy of the spring, and work done by friction. There is uncertainty about the calculations leading to different results, particularly regarding the expected speed of the disc.

Discussion Status

Some participants are verifying their calculations and questioning the correctness of the expected answer provided in the review materials. There is a mix of attempts to clarify the problem setup and check arithmetic, with no consensus reached on the correct speed of the disc.

Contextual Notes

Participants express frustration over the perceived difficulty of the problem and the lack of immediate assistance, indicating a broader context of similar problems being discussed in the forum. There is mention of an upcoming physics exam, adding urgency to the discussion.

salman213
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1. A toy gun fires a 9.41-g projectile disc by using a compressed sprint (k=1.72 x 10^3 N/m) and a 13.1 cm long barrel. As the disc travvels through the barrel it experiences a constant frictional force of 0.13 N. If the spring is compressed 14 mm what is the speed of the disc as it leaves the gun?



2. Ek = 1/2mv^2
Ee = 1/2kx^2
W = F*d




3. DeltaEe - W = DeltaEk ?

1/2kx^2 - F*d = 1/2mv^2
1/2(1.72 x 10^3)(0.014)^2 - (0.13*0.131) = 1/2(0.00941)(v)^2
v = 5.67 m/s BUT THATS WRONG, I have a physics exam some day this week and i was just reviewin gmy notes and trying old review problems he gave us..and on the review it says the answer to this question is 1.8 m/s...i don't get what I am doing wrong :(
 
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im not what's wrong but after going through it again if i make v = 1.8m/s as it says the answer is then the Ek part equals 0.015 while the left equals 0.15 so maybe imd oing a conversion mistake but i chekced i don't see where :S:S:S
 
hmm come on guys I am sure someones knows what I am doing wrong this is grade 12physics :S :S :S
 
Is it that hard? or no one wants to help lool..
 
salman213 said:
Is it that hard? or no one wants to help lool..

there has been many many problems of this kind posting during these weeks. We who hang around on this forum have studies to do ourselves and can not help everyone all the time.

1. find out how much potential energy there is in the spring when pulled back in the initial position.

2. All this energy is then reformed as kinetic energy + friction work.

3. Find out the work done by the friction, if the force is given and the lenght, what is then the work done by the friction.

4. The energy remained is kinetic energy of bullet.
 
2. Ek = 1/2mv^2
Ee = 1/2kx^2
W = F*d



3. DeltaEe = DeltaEk + W

1/2kx^2 - F*d = 1/2mv^2

1/2(1.72 x 10^3)(0.014)^2 - (0.13*0.131) = 1/2(0.00941)(v)^2
v = 5.67 m/s

Does that look right to you? cause that's what i did and i don't get 1.8m/s which is supposedly the right answer
 
I checked your arithmetic. I also seem to think that your equation is right. I would take it up with the prof. It's possible his answer to the review problem was wrong.
 
i get the same lol
 
alright thanks! :)
 

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