The launching mechanism of a toy gun consists of a spring of unknown spring constant,

If the spring is compressed a distance of 0.180 m and the gun fired vertically as shown, the gun can launch a 18.0-g projectile from rest to a maximum height of 16.0 m above the starting point of the projectile.

(a) Neglecting all resistive forces, describe the mechanical energy transformations that occur from the time the gun is fired until the projectile reaches its maximum height.

(b) Neglecting all resistive forces, determine the spring constant. (N/m)


(c) Neglecting all resistive forces, find the speed of the projectile as it moves through the equilibrium position of the spring (where x = 0) (m/s)

(a) The elastic (or spring potential) energy in the spring is transformed to kinetic energy when it’s released. Then, this kinetic energy changes entirely to gravitational potential energy when the projectile comes to a stop momentarily at the top of its flight path.

(b )So from (a), we can apply the law of conservation of mechanical energy and equate the initial spring potential energy to the gravitational potential energy of the projectile and solve for k:

0.5kx² = mgh
k = 2mgh/x²
= 2(0.0180kg)(9.81m/s²)(16.0m) / (0.180m)²
= 174N/m

(c) Now equate the spring potential energy to the kinetic energy and solve for v:

0.5kx² = 0.5mv²
v = √[kx²/m]
= √[(174N/m)(0.180m) / 0.0180kg]
= 17.7m/s

Hope this helps.