Calculating Muzzle Velocity of a Spring-Powered Civil War Cannon

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The discussion focuses on calculating the muzzle velocity of a spring-powered Civil War cannon designed for reenactments. The cannon uses a spring with a constant of 3000 N/m to launch a 1.0 kg plastic ball, requiring a minimum speed of 2.0 m/s for the latch to engage after compressing the spring by 30 cm. The minimum force needed to compress and latch the spring is calculated to be 459 N, considering friction and other forces. The muzzle velocity of the ball, after traveling a total distance of 1.5 m, is determined to be approximately 16.5 m/s, assuming the cannon is parallel to the ground. The discussion also touches on the concepts of kinetic and potential energy in the context of the cannon's operation.
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As a hobby, you like to participate in reenactments of Civil War battles. Civil War cannons were muzzle loeaded, meaning that the gunpowder and the cannonball were inserted into the output end of the muzzle, then tamped into place with a long plunger. To recreate the authenticity of muzzle-loaded cannons, but without the danger of real cannons, Civil War buffs have invented a spring-powered cannon that fires a 1.0kg plastic ball. A spring, with constant 3000N/m, is mounted at the back of the barrel. You place a ball in the barrel, then use a long plunger to press the ball against the spring and lock the spring into place, ready for firing. In order for the latch to catch, the ball has to be moving at a speed of at least 2.0m/s when the spring has been compressed 30cm. The coefficient of friction of the ball in the barrel is 0.30. The plunger doens't touch the sides of the barrel.

a) if you push the plunger with a constant force, what is the minimum force that you must use to compress and latch the spring? You can assume that no effort was required to push the ball down the barrel to where it first contacts the spring.

b) what is the cannon's muzzle velocity if the ball travels a total distance of 1.5m to the end of the barrel?
 
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For a did this this
Fy=0=Fn-mg
Ffr=(mu)mg=2.94N
delta(K)=Wc+Wdiss+Wext
2J=-135J-0.882J+Wext
Wext=137.87J
W=Fd
F=459N
 
for b I'm a little bit confused I believe that:
delta(K)=Delta(U)+Wdiss
so velocity as soon as it leaves the spring is
vf=16.5m/s
 
I believe you need the angle of elevation of the cannon.
 
hmm... i think you just assume it is parallel to the ground
 
What is U and K in your equation?
 
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