Question

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?

Answer 1

Answer:

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

Answer 2

Answer:

Explanation:

mass, m = 5.9 g = 0.0059 kg

Spring constant, K = 8 N/m

length of barrel, s = 17 cm = 0.17 m

frictional force, f = 0.035 N

Compression in spring, Δs = 5.7 cm = 0.057 m

let the speed of the projectile is v.

Energy stored in the spring, E = 0.5 KΔs²

E = 0.5 x 8 x 0.057 x 0.057 = 0.012996 J

Work done by the friction force, W = f x s = 0.035 x 0.17 = 0.00595 J

Energy used to move the projectile, E' = E - W

E' = 0.012996 - 0.00595 = 0.007046 J

E' = 1/2 mv²

0.007046 = 0.5 x 0.0059 x v²

v² = 2.3885

v = 1.55 m/s