A toy gun uses a spring to project a 5.9-g soft rubber sphere horizontally. The spring constant is 8.0 N/m, the barrel of the gun is 17 cm long, and a constant frictional force of 0.035 N exists between barrel and projectile. With what speed does the projectile leave the barrel if the spring was compressed 5.7 cm for this launch?

Explanation:

Stored energy in spring = 1/2 k x² , k is spring constant , x is compression.

= 1/2 x 8 x (5.7 x 10⁻²)²

= 129.96 x 10⁻⁴ J

Energy lost due to friction = force x distance

= .035 x .17

= .00595 J

Energy used in providing kinetic energy to projectile.  

129.96 x 10⁻⁴  - .00595

.012996 - .00595

= .007046 J

So

1/2 m v² = .007046

v² = .007046  x 2 / .0059

= 2.3885

v = 1.545 m /s

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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this is poiselle's equation from fluid mechanic in physics. mech eng