Answer to Question #85830 in Quantum Mechanics for Julie

Question

The massless spring of a spring gun has a force constant k = 0.04 N/cm. When the gun is aimed vertically, a 30 g projectile is shot to a height of 5.0 m above the end of the expanded spring. How much was the spring compressed initially? (Find d, in meters.)

Answer 1

In accordance with the law of conservation of energy kd²/2 = mgh , then d = (2mgh/k)^1/2 = (2*0.3*5/4)^1/2 = 0.87 (m).

d = 0.87 m