How Is Initial Velocity Calculated in a Spring Gun?

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To calculate the initial velocity of a spring gun, the spring's potential energy must be equated to the kinetic energy of the ball. The spring constant is 2.0 N/cm, and with a compression of 3.0 cm, the potential energy can be calculated using the formula E = 0.5kx^2. This energy is converted to kinetic energy at the moment of release, represented by the equation E(k) = 0.5mv^2. The mass of the ball is 0.15 kg, allowing for the calculation of initial velocity by solving the equation after substituting the known values. The gravitational potential energy is not needed in this scenario since no height is specified.
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A spring gun has a spring with a constant of 2.0 N/cm. If the spring is compressed 3.0cm and the gun is loaded with a .15 kg ball, what will be the initial velocity?



Equations such as E(k subscript)=0.5mv^2
E(p subscript)-mgh



3. i think that since kinetic energy is equall to potenial, i used the equation 0.5(m)(v)^2=mgh.
So the (m) crosses of from both sides and i solved for v^2.
Yeah that's as far as i got and I am confused as to what to do next or what to cross off next?
 
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Kinetic energy of the ball is equal to potential energy of the spring, so:
mv^2/2=kx^2.
You don't need potential energy of the ball here.
 
In this case, since the gun is told to be at a certain height, all the energy is potential, but its spring potential energy. Use E=(.5)kx^2 with k being the constant and x being the distance the spring is compressed. All this energy becomes linetic at the point of release so set it equal to the kinetic energy equation and solve for v using the given m.

If the gun was at a given height, then you would use the gravitational potential energy equation in the problem somehow, but since no height is given, you just use the ones I told you.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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