Energy Conservation of a toy gun

AI Thread Summary
A toy gun uses a compressed spring to fire a projectile disc, and the discussion revolves around calculating the disc's speed as it exits the barrel. The user initially calculates the speed to be 5.67 m/s using energy conservation equations, but notes that a review problem states the correct answer is 1.8 m/s. The conversation highlights the importance of accounting for potential energy in the spring, work done by friction, and the relationship between kinetic and potential energy. Users suggest verifying calculations and considering the possibility of an error in the review problem's answer. The discussion emphasizes the challenges of understanding energy conservation in physics problems.
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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