Can the gun fire the mass if the gun is fired straight up?

Question

Suppose a 0.30-kg mass on a spring-loaded gun that has been compressed 0.10m has elastic potential energy of 1.0 J. How high above the spring's equilibrium point, in meters, can the gun fire the mass if the gun is fired straight up?Calculate to 2 decimal points.

Answer 1

Total energy while compressedTE = PE = 1//2 k dX^2 = 1 kg-(m/sec)^2; where dX = .1 m

Total energy upon leaving the decompressed spring TE = KE = 1/2 mV^2; where m = .3 kg and V is the launch velocity.

From the consevation of energy total compressed energy PE = 1 = 1/2 mV^2 = mgH = PE the potential energy after the kinetic energy is expended where g = 9.81 m/sec^2 and H = ? the height you are looking for; so that H = 1/mg = 1/(.3*9.81) = .339 meter.  This answer discounts friction losses while the spring decompresses and drag losses while the mass is lifting to H.

The physics is the conservation of energy.  The total energy while compressed, when decompressed, and max height are equal to each other as no friction or drag losses are assumed.

PS:  The equilibrium point of a spring is not the same as its mid-point.  The equilibrium point is when dX = 0, there is no compression or stretching, so there is no force on the spring.  The mid point is L/2 where L is the physical length of the spring when dX = 0.  Thus, when finding the height above the spring's equilibrium, we are looking for the height above the end of the spring at L, not above L/2.  As we don't know L, we cannot find L/2.

Answer 2

Yes.

.24 m

You have .34 m of potential energy at the compressed spring.  It will decompress .1 m (raising the weight that height).  So the total height above the midpoint of the spring will be:

.34 m - .1 m = .24 m