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

0.11 m/s and 0.28 m/s should be correct

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.