A vertically oriented \"cannon\" consists of a tube with a spring that extends a
ID: 1337680 • Letter: A
Question
A vertically oriented "cannon" consists of a tube with a spring that extends all the way from the bottom to the top of the tube when not compressed. The "cannon ball" is a marble of mass, m= 0.1kg. The spring's spring constant is k=5 N/m. A F=10N is used to compress the spring by a distance d=2.0 m, at which time it is locked in the compressed statem and the marble is loaded. Then the lock is released, so the marble is propelled out of the tube. How far above the top of the tube does the marble rise? Call that distance, h. Use coordinate system in which the +y direction is up, and the marble's loaded position is y=0. Neglect the diameter of the marble.
Explanation / Answer
Initially ,
Spring Potential energy = kx^2 /2 = 5 x 2^2 /2 = 10 J
Gravitational PE = 0
KE = 0
Finally (At highest position),
SPE = 0
GPE = 0.1 x 9.81 x (2 +h) = 0.981(2 + h)
KE = 0
Using energy conservation,
10 + 0 + 0 = 0 + 0.981(2 +h) + 0
h = 8.19 m